Solution

\[ \color{magenta}\mathbf{(a + b)^2 = a^2 + 2ab + b^2} \]

(i) \( (7x + 4y)^2 \)

\[ \begin{aligned} & a = 7x \ , \ b = 4y \\[8pt] & = (7x + 4y)^2 \\[6pt] &= (7x)^2 + 2(7x)(4y) + (4y)^2 \\[6pt] &= \color{red} 49x^2 + 56xy + 16y^2 \end{aligned} \]

(ii) \( \left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2 \)

\[ \begin{aligned} & a = \frac{7}{5}x \ , \ b = \frac{3}{2}y \\[8pt] & = \left(\frac{7}{5}x + \frac{3}{2}y\right)^2 \\[6pt] &= \left(\frac{7}{5}x\right)^2 + \cancel2\left(\frac{7}{5}x\right)\left(\frac{3}{\cancel2}y\right) + \left(\frac{3}{2}y\right)^2 \\[8pt] &= \color{red} \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2 \end{aligned} \]

(iii) \( (2.5p + 1.5q)^2 \)

\[ \begin{aligned} & a = 2.5p \ , \ b = 1.5q \\[8pt] & =(2.5p + 1.5q)^2 \\[6pt] &= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2 \\[6pt] &= 6.25p^2 + 2(3.75pq) + 2.25q^2 \\[6pt] &= \color{red}\mathbf{6.25p^2 + 7.5pq + 2.25q^2} \end{aligned} \]

(iv) \( \left(\dfrac{3}{4}s + 8t\right)^2 \)

\[ \begin{aligned} & a = \frac{3}{4}s \ , \ b = 8t \\[8pt] & = \left(\frac{3}{4}s + 8t\right)^2 \\[6pt] &= \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2 \\[8pt] &= \frac{9}{16}s^2 + 2\left(\frac{3}{\cancel4}s\right)(\cancel{8}^2t) + 64t^2 \\[8pt] &= \color{red}\mathbf{\frac{9}{16}s^2 + 12st + 64t^2} \end{aligned} \]

(v) \( \left(x + \dfrac{1}{2y}\right)^2 \)

\[ \begin{aligned} & a = x \ , \ b = \frac{1}{2y} \\[8pt] & = \left(x + \frac{1}{2y}\right)^2 \\[6pt] & = (x)^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2 \\[8pt] &= x^2 + \cancel{2}x\left(\frac{1}{\cancel{2}y}\right) + \frac{1}{4y^2} \\[8pt] &= \color{red} x^2 + \frac{x}{y} + \frac{1}{4y^2} \end{aligned} \]

(vi) \( \left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2 \)

\[ \begin{aligned} & a = \frac{1}{x} \ , \ b = \frac{1}{y} \\[8pt] & = \left(\frac{1}{x} + \frac{1}{y}\right)^2 \\[6pt] & = \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2 \\[8pt] &= \color{red}\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2} \end{aligned} \]

Solution

\[ \color{magenta}\mathbf{(a + b)^2 = a^2 + 2ab + b^2} \]

(i) \( (64)^2 \)

\[ \begin{aligned} a & = 60 \ , \ b = 4 \\[8pt] (64)^2 &= (60 + 4)^2 \\[6pt] &= (60)^2 + 2(60)(4) + (4)^2 \\[6pt] &= 3600 + 480 + 16 \\[6pt] &= \color{red}\mathbf{4096} \end{aligned} \]

(ii) \( (105)^2 \)

\[ \begin{aligned} a & = 100 \ , \ b = 5 \\[8pt] (105)^2 & = (100 + 5)^2 \\[6pt] &= (100)^2 + 2(100)(5) + (5)^2 \\[6pt] &= 10000 + 1000 + 25 \\[6pt] &= \color{red}\mathbf{11025} \end{aligned} \]

(iii) \( (205)^2 \)

\[ \begin{aligned} a & = 200 \ , \ b = 5 \\[8pt] (205)^2 &= (200 + 5)^2 \\[6pt] &= (200)^2 + 2(200)(5) + (5)^2 \\[6pt] &= 40000 + 2000 + 25 \\[6pt] &= \color{red}\mathbf{42025} \end{aligned} \]

Solution

\[ \color{magenta}\mathbf{a^2 + 2ab + b^2 = (a + b)^2} \]

(i) \( 9x^2 + 24xy + 16y^2 \)

\[ \begin{aligned} 9x^2 &= (3x)^2 \\[6pt] 16y^2 &= (4y)^2 \\[6pt] 24xy &= 2(3x)(4y) \\[8pt] & = 9x^2 + 24xy + 16y^2 \\[6pt] &= (3x)^2 + 2(3x)(4y) + (4y)^2 \\[6pt] &= \color{red} {(3x + 4y)^2} \\[6pt] & = \color{green} {(3x + 4y)(3x + 4y)} \end{aligned} \]

(ii) \( 4s^2 + 20st + 25t^2 \)

\[ \begin{aligned} 4s^2 &= (2s)^2 \\[6pt] 25t^2 &= (5t)^2 \\[6pt] 20st &= 2(2s)(5t) \\\\ & = 4s^2 + 20st + 25t^2 \\[6pt] &= (2s)^2 + 2(2s)(5t) + (5t)^2 \\[6pt] &= \color{red}\mathbf{(2s + 5t)^2} \\[6pt] & = \color{green}\mathbf{(2s + 5t)(2s + 5t)} \end{aligned} \]

(iii) \( 49x^2 + 28xy + 4y^2 \)

\[ \begin{aligned} 49x^2 &= (7x)^2 \\[6pt] 4y^2 &= (2y)^2 \\[6pt] 28xy &= 2(7x)(2y) \\\\[8pt] & = 49x^2 + 28xy + 4y^2 \\[6pt] &= (7x)^2 + 2(7x)(2y) + (2y)^2 \\[6pt] &= \color{red} {(7x + 2y)^2} \\[6pt] & = \color{green} {(7x + 2y)(7x + 2y)} \end{aligned} \]

(iv) \( 64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2 \)

\[ \begin{aligned} 64p^2 &= (8p)^2 \\[6pt] \frac{4}{9}q^2 &= \left(\frac{2}{3}q\right)^2 \\[8pt] \frac{32}{3}pq &= 2(8p)\left(\frac{2}{3}q\right) \\\\[8pt] & = 64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 \\[6pt] &= (8p)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2 \\[8pt] &= {\color{red}\mathbf{\left(8p + \frac{2}{3}q\right)^2}} \quad \\[8pt] & = \color{green}\left( \mathbf{8p + \frac{2}{3}q}\right)\left(\mathbf{8p + \frac{2}{3}q}\right) \end{aligned} \]

*(v) \( 3a^2 + 4ab + \dfrac{4}{3}b^2 \)

\[ \begin{aligned} \color{magenta}\textbf{Taking out } \frac{1}{3} & \color{magenta} \textbf{ as a common factor:} \\[6pt] 3a^2 + 4ab + \frac{4}{3}b^2 &= \frac{1}{3}\left(9a^2 + 12ab + 4b^2\right) \\[8pt] 9a^2 &= (3a)^2 \\[6pt] 4b^2 &= (2b)^2 \\[6pt] 12ab &= 2(3a)(2b) \\\\[8pt] \implies \frac{1}{3}\left(9a^2 + 12ab + 4b^2\right) &= \frac{1}{3}\left[(3a)^2 + 2(3a)(2b) + (2b)^2\right] \\[8pt] &= \color{red}\mathbf{\frac{1}{3}(3a + 2b)^2} \\[6pt] & = \color{green} \frac{1}{3}(3a + 2b) (3a + 2b) \end{aligned} \]

*(vi) \( \dfrac{9}{5}s^2 + 6sv + 5v^2 \)

\[ \begin{aligned} \color{magenta}\textbf{Taking out } \frac{1}{5} & \color{magenta} \textbf{ as a common factor:} \\[6pt] \frac{9}{5}s^2 + 6sv + 5v^2 &= \frac{1}{5}\left(9s^2 + 30sv + 25v^2\right) \\\\[8pt] 9s^2 &= (3s)^2 \\[6pt] 25v^2 &= (5v)^2 \\[6pt] 30sv &= 2(3s)(5v) \\\\[8pt] \implies \frac{1}{5}\left(9s^2 + 30sv + 25v^2\right) &= \frac{1}{5}\left[(3s)^2 + 2(3s)(5v) + (5v)^2\right] \\[8pt] &= \color{red}\mathbf{\frac{1}{5}(3s + 5v)^2} \\[6pt] & = \color{green} \mathbf{ \frac{1}{5}(3s + 5v) (3s + 5v)} \end{aligned} \]

Solution

\[ \color{magenta}\mathbf{(a - b)^2 = a^2 - 2ab + b^2} \]

(i) \( (79)^2 \)

\[ \begin{aligned} a & = 80 \ , \ b = 1 \\[8pt] (79)^2 & = (80 - 1)^2 \\[6pt] &= (80)^2 - 2(80)(1) + (1)^2 \\[6pt] &= 6400 - 160 + 1 \\[6pt] &= 6240 + 1 \\[6pt] &= \color{red}\mathbf{6241} \end{aligned} \]

(ii) \( (193)^2 \)

\[ \begin{aligned} a &= 200 \ , \ b = 7 \\[8pt] (193)^2 &= (200 - 7)^2 \\[6pt] &= (200)^2 - 2(200)(7) + (7)^2 \\[6pt] &= 40000 - 2800 + 49 \\[6pt] &= 37200 + 49 \\[6pt] &= \color{red}\mathbf{37249} \end{aligned} \]

(iii) \( (299)^2 \)

\[ \begin{aligned} a & = 300 \ , \ b = 1 \\[8pt] (299)^2 & = (300 - 1)^2 \\[6pt] &= (300)^2 - 2(300)(1) + (1)^2 \\[6pt] &= 90000 - 600 + 1 \\[6pt] &= 89400 + 1 \\[6pt] &= \color{red}\mathbf{89401} \end{aligned} \]

Solution

(i) \( 117^2 \)

\[ \begin{aligned} 117^2 &= (100 + 10 + 7)^2 \\[6pt] &= (100)^2 + (10)^2 + (7)^2 + 2(100)(10) + 2(10)(7) + 2(100)(7) \\[6pt] &= 10000 + 100 + 49 + 2000 + 140 + 1400 \\[6pt] &= \color{red}\mathbf{13689} \end{aligned} \]

(ii) \( 78^2 \)

\[ \begin{aligned} 78^2 &= (80 - 2)^2 \\[6pt] &= (80)^2 - 2(80)(2) + (2)^2 \\[6pt] &= 6400 - 320 + 4 \\[6pt] &= 6080 + 4 \\[6pt] & = \color{red}\mathbf{6084} \end{aligned} \]

(iii) \( 198^2 \)

\[ \begin{aligned} 198^2 &= (200 - 2)^2 \\[6pt] &= (200)^2 - 2(200)(2) + (2)^2 \\[6pt] &= 40000 - 800 + 4 \\[6pt] &= 39200 + 4 \\[6pt] & = \color{red}\mathbf{39204} \end{aligned} \]

(iv) \( 214^2 \)

\[ \begin{aligned} 214^2 &= (200 + 10 + 4)^2 \\[6pt] &= (200)^2 + (10)^2 + (4)^2 + 2(200)(10) + 2(10)(4) + 2(200)(4) \\[6pt] &= 40000 + 100 + 16 + 4000 + 80 + 1600 \\[6pt] &= \color{red}\mathbf{45796} \end{aligned} \]

(v) \( 1104^2 \)

\[ \begin{aligned} 1104^2 &= (1100 + 4)^2 \\[6pt] &= (1100)^2 + 2(1100)(4) + (4)^2 \\[6pt] &= 1210000 + 8800 + 16 \\[6pt] &= \color{red}\mathbf{1218816} \end{aligned} \]

(vi) \( 1120^2 \)

\[ \begin{aligned} 1120^2 &= (1100 + 20)^2 \\[6pt] &= (1100)^2 + 2(1100)(20) + (20)^2 \\[6pt] &= 1210000 + 44000 + 400 \\[6pt] &= \color{red}\mathbf{1254400} \end{aligned} \]

Solution

(i) \( 16y^2 - 24y + 9 \)

\[ \begin{aligned} & \color{magenta} a^2 - 2ab + b^2 \implies (a - b)^2 \\[6pt] & = 16y^2 - 24y + 9 \\[6pt] &= (4y)^2 - 2(4y)(3) + (3)^2 \\[6pt] &= \color{red}\mathbf{(4y - 3)^2} \end{aligned} \]

(ii) \( \dfrac{9}{4}s^2 + 6st + 4t^2 \)

\[ \begin{aligned} & \color{magenta} a^2 + 2ab + b^2 \implies (a + b)^2 \\[8pt] & = \frac{9}{4}s^2 + 6st + 4t^2 \\[8pt] &= \left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2 \\[8pt] &= \color{red}\mathbf{\left(\frac{3}{2}s + 2t\right)^2} \end{aligned} \]

(iii) \( \dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2 \)

Rearranging terms in standard form: \[ \frac{m^2}{9} + \frac{k^2}{4} + 9n^2 + \frac{mk}{3} + 3nk + 2mn \] \[ \color{magenta} a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \implies (a + b + c)^2 \] \[ \begin{aligned} &= \left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) + 2\left(\frac{k}{2}\right)(3n) + 2\left(\frac{m}{3}\right)(3n) \\[8pt] &= \color{red}\mathbf{\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2} \end{aligned} \]

(iv) \( \dfrac{p^2}{16} - 2 + \dfrac{16}{p^2} \)

\[ \begin{aligned} & \color{magenta} a^2 - 2ab + b^2 \implies (a - b)^2 \\[8pt] & = \frac{p^2}{16} - 2 + \frac{16}{p^2} \\[8pt] &= \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2 \\[8pt] &= \color{red}\mathbf{\left(\frac{p}{4} - \frac{4}{p}\right)^2} \end{aligned} \]

(v) \( 9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc \)

Notice that only terms containing \( b \) are negative (\( -12ab \) and \( -4bc \)): \[ \begin{aligned} & = 9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc \\[8pt] &= (3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(3a)(c) \\[6pt] &= \color{red}\mathbf{(3a - 2b + c)^2} \end{aligned} \]

Solution

(i) \( (p + 3q + 7r)^2 \)

\[ \begin{aligned} & = (p + 3q + 7r)^2 \\[6pt] &= (p)^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p) \\[6pt] &= \color{red}\mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr} \end{aligned} \]

(ii) \( (3x - 2y + 4z)^2 \)

\[ \begin{aligned} & = (3x - 2y + 4z)^2 \\[6pt] &= [3x + (-2y) + 4z]^2 \\[6pt] &= (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) \\[6pt] &= \color{red}\mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx} \end{aligned} \]

Solution

\[ \begin{aligned} {\color{magenta}\textbf{L.H.S.:}} & \\[6pt] (a + b - c)^2 &= a^2 + b^2 + c^2 + 2ab - 2bc - 2ca \\[6pt] (a - b + c)^2 &= a^2 + b^2 + c^2 - 2ab - 2bc + 2ca \\[6pt] (a - b - c)^2 &= a^2 + b^2 + c^2 - 2ab + 2bc - 2ca \\\\ \text{L.H.S.} &= 3a^2 + 3b^2 + 3c^2 + (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ca + 2ca - 2ca) \\[8pt] &= \color{blue}\mathbf{3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca} \\[12pt] \text{R.H.S.} &= 2a^2 + 2b^2 + 2c^2 \\[8pt] \implies \text{L.H.S.} &\ne \text{R.H.S.} \end{aligned} \]

Answer No, it is not an identity because \(\text{L.H.S.} \ne \text{R.H.S.}\) (the expanded sum is \( 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca \)).

Solution

(i) \( s^2 - 11s + 24 \)

\[ \begin{aligned} \color{magenta} x^2 + (a + b)x + ab & = \color{magenta} (x + a)(x + b) \\[6pt] \text{Product } (ab) &= 24, \quad \text{Sum } (a + b) = -11 \\[6pt] \implies a &= -3, \quad b = -8 \\[6pt] & = s^2 - 11s + 24 \\[6pt] &= s^2 - 3s - 8s + 24 \\[6pt] &= s(s - 3) - 8(s - 3) \\[6pt] &= \color{red}\mathbf{(s - 3)(s - 8)} \end{aligned} \]

(ii) \( (\underline{\hspace{1.2cm}})(x + 1) = (3x^2 - 4x - 7) \)

\[ \begin{aligned} & = 3x^2 - 4x - 7 \\[6pt] &= 3x^2 - 7x + 3x - 7 \\[6pt] &= 3x^2 + 3x - 7x - 7 \\[6pt] &= 3x(x + 1) - 7(x + 1) \\[6pt] &= \color{red}\mathbf{(3x - 7)}\mathbf{(x + 1)} \end{aligned} \]

(iii) \( 10x^2 - 11x - 6 = (2x - \underline{\hspace{0.8cm}})(\underline{\hspace{0.8cm}} + 2) \)

\[ \begin{aligned} & = 10x^2 - 11x - 6 \\[6pt] &= 10x^2 - 15x + 4x - 6 \\[6pt] &= 5x(2x - 3) + 2(2x - 3) \\[6pt] &= (2x - 3)(5x + 2) \\\\[6pt] \implies 10x^2 - 11x - 6 &= (2x - \color{red}\mathbf{3})(\color{red}\mathbf{5x} + 2) \end{aligned} \]

(iv) \( 6x^2 + 7x + 2 = (\underline{\hspace{1.2cm}})(\underline{\hspace{1.2cm}}) \)

\[ \begin{aligned} \text{Product} = 6 \times 2 &= 12, \quad \text{Sum} = 7 \implies 4 \text{ and } 3 \\[6pt] & = 6x^2 + 7x + 2 \\[6pt] &= 6x^2 + 4x + 3x + 2 \\[6pt] &= 2x(3x + 2) + 1(3x + 2) \\[6pt] &= \color{red}\mathbf{(2x + 1)(3x + 2)} \end{aligned} \]

Solution

(i) \( (41)^2 \)

\[ \begin{aligned} \color{magenta}(a + b)^2 &= \color{magenta} a^2 + 2ab + b^2 \\[6pt] (41)^2 &= (40 + 1)^2 \\[6pt] &= (40)^2 + 2(40)(1) + (1)^2 \\[6pt] &= 1600 + 80 + 1 \\[6pt] &= \color{red}\mathbf{1681} \end{aligned} \]

(ii) \( (27)^2 \)

\[ \begin{aligned} \color{magenta} (a - b)^2 &= \color{magenta} a^2 - 2ab + b^2 \\[6pt] (27)^2 &= (30 - 3)^2 \\[6pt] &= (30)^2 - 2(30)(3) + (3)^2 \\[6pt] &= 900 - 180 + 9 \\[6pt] &= 720 + 9 \\[6pt] &= \color{red}\mathbf{729} \end{aligned} \]

(iii) \( (23 \times 17) \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) & = \color{magenta} a^2 - b^2 \\[6pt] 23 \times 17 &= (20 + 3)(20 - 3) \\[6pt] &= (20)^2 - (3)^2 \\[6pt] &= 400 - 9 \\[6pt] &= \color{red}\mathbf{391} \end{aligned} \]

(iv) \( (135)^2 \)

\[ \begin{aligned} \color{magenta}(a - b)^2 &= \color{magenta} a^2 - 2ab + b^2 \\[6pt] (135)^2 &= (140 - 5)^2 \\[6pt] &= (140)^2 - 2(140)(5) + (5)^2 \\[6pt] &= 19600 - 1400 + 25 \\[6pt] &= 18200 + 25 \\[6pt] &= \color{red}\mathbf{18225} \end{aligned} \]

(v) \( (97)^2 \)

\[ \begin{aligned} \color{magenta} (a - b)^2 &= \color{magenta}a^2 - 2ab + b^2 \\[6pt] (97)^2 &= (100 - 3)^2 \\[6pt] &= (100)^2 - 2(100)(3) + (3)^2 \\[6pt] &= 10000 - 600 + 9 \\[6pt] &= 9400 + 9 \\[6pt] &= \color{red}\mathbf{9409} \end{aligned} \]

(vi) \( (18 \times 29) \)

\[ \begin{aligned} \color{magenta} (x + a)(x + b) &= \color{magenta}x^2 + (a + b)x + ab \\[6pt] 18 \times 29 &= (20 - 2)(20 + 9) \\[6pt] &= (20)^2 + (-2 + 9)(20) + (-2)(9) \\[6pt] &= 400 + 7(20) - 18 \\[6pt] &= 400 + 140 - 18 \\[6pt] &= 540 - 18 \\[6pt] &= \color{red}\mathbf{522} \end{aligned} \]

(vii) \( (34 \times 43) \)

\[ \begin{aligned} \color{magenta} (x + a)(x + b) &= \color{magenta} x^2 + (a + b)x + ab \\[6pt] 34 \times 43 &= (40 - 6)(40 + 3) \\[6pt] &= (40)^2 + (-6 + 3)(40) + (-6)(3) \\[6pt] &= 1600 + (-3)(40) - 18 \\[6pt] &= 1600 - 120 - 18 \\[6pt] &= 1480 - 18 \\[6pt] &= \color{red}\mathbf{1462} \end{aligned} \]

(viii) \( (205)^2 \)

\[ \begin{aligned} \color{magenta}(a + b)^2 & = \color{magenta} a^2 + 2ab + b^2 \\[6pt] (205)^2 & = (200 + 5)^2 \\[6pt] &= (200)^2 + 2(200)(5) + (5)^2 \\[6pt] &= 40000 + 2000 + 25 \\[6pt] &= \color{red}\mathbf{42025} \end{aligned} \]

Solution

(i) \( 9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc \)

\[ \color{magenta} {x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2} \] \[ \begin{aligned} & \text{Terms containing } b \text{ are negative } (-6ab \text{ and } -4bc) \\[6pt] &= 9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc \\[6pt] &= (3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(3a)(2c) \\[6pt] &= \color{red}\mathbf{(3a - b + 2c)^2} \\[6pt] &= \color{green}\mathbf{(3a - b + 2c)(3a - b + 2c)} \end{aligned} \]

(ii) \( 16s^2 + 25t^2 - 40st \)

\[ \begin{aligned} \color{magenta}a^2 - 2ab + b^2 &= \color{magenta}(a - b)^2 \\[6pt] &= 16s^2 - 40st + 25t^2 \\[6pt] &= (4s)^2 - 2(4s)(5t) + (5t)^2 \\[6pt] &= \color{red}\mathbf{(4s - 5t)^2} \\[6pt] &= \color{green}\mathbf{(4s - 5t)(4s - 5t)} \end{aligned} \]

(iii) \( r^2 - r - 42 \)

\[ \begin{aligned} \color{magenta} x^2 + (a + b)x + ab &= \color{magenta}(x + a)(x + b) \\[6pt] \text{Product } (ab) &= -42, \quad \text{Sum } (a + b) = -1 \\[6pt] \implies a &= -7, \quad b = 6 \\[6pt] & = r^2 - r - 42 \\[6pt] &= r^2 - 7r + 6r - 42 \\[6pt] &= r(r - 7) + 6(r - 7) \\[6pt] &= \color{red}\mathbf{(r - 7)(r + 6)} \end{aligned} \]

(iv) \( 49g^2 + 14gh + h^2 \)

\[ \begin{aligned} & \color{magenta}\mathbf{a^2 + 2ab + b^2 = (a + b)^2} \\[6pt] & = 49g^2 + 14gh + h^2 \\[6pt] &= (7g)^2 + 2(7g)(h) + (h)^2 \\[6pt] &= \color{red}\mathbf{(7g + h)^2} \\[6pt] &= \color{green}\mathbf{(7g + h)(7g + h)} \end{aligned} \]

(v) \( 64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw \)

\[ \begin{aligned} & \color{magenta}{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2} \\[8pt] & = 64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw \\[8pt] & \text{Terms containing u are negative } (-176uv \text{ and } -32uw) \\[4pt] & = (-8u)^2 + (11v)^2 + (2w)^2 + 2(-8u)(11v) + 2(11v)(2w) + 2(-8u)(2w) \\[6pt] &= \color{red}\mathbf{(-8u + 11v + 2w)^2} \quad \left(\text{or } \color{red}\mathbf{(8u - 11v - 2w)^2}\right) \\[6pt] &= \color{green}\mathbf{(8u - 11v - 2w)(8u - 11v - 2w)} \end{aligned} \]

Solution

(i) \( \dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} \)

\[ \begin{aligned} & \color{magenta}\textbf{ Numerator} \\[6pt] & = 3p^2 - 3pq - 18q^2 \\[6pt] &= 3(p^2 - pq - 6q^2) \\[6pt] &= 3[p^2 - 3pq + 2pq - 6q^2] \\[6pt] &= 3[p(p - 3q) + 2q(p - 3q)] \\[6pt] &= \mathbf{3(p - 3q)(p + 2q)} \\[10pt] & \color{magenta}\textbf{ Denominator} \\[6pt] & = p^2 + 3pq - 10q^2 \\[6pt] &= p^2 + 5pq - 2pq - 10q^2 \\[6pt] &= p(p + 5q) - 2q(p + 5q) \\[6pt] &= \mathbf{(p + 5q)(p - 2q)}\\[10pt] \frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} &= \color{red}\mathbf{\frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}} \end{aligned} \]

(ii) \( \dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} \)

\[ \begin{aligned} &\color{magenta}\textbf{Numerator} \\[6pt] \color{magenta} (a - b)^3 &= \color{magenta}a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] & = n^3 - 3n^2m + 3nm^2 - m^3 \\[6pt] &= (n - m)^3 \\\\[10pt] &\color{magenta}\textbf{Denominator} \\[6pt] \color{magenta}(a - b)^2 & = \color{magenta}a^2 - 2ab + b^2 \\[6pt] & = 5m^2 - 10mn + 5n^2 \\[6pt] &= 5(m^2 - 2mn + n^2) \\[6pt] &= 5(m - n)^2 \\[6pt] &= 5(n - m)^2 \\\\[10pt] \frac{(n - m)^3}{5(n - m)^2} &= \frac{(n - m)\cancel{(n - m)^2}}{5\cancel{(n - m)^2}} \\[8pt] &= \color{red}\mathbf{\frac{n - m}{5}} \end{aligned} \]

(iii) \( \dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] \color{magenta} a^3 + b^3 + c^3 - 3abc &= \color{magenta} (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\[6pt] a = w, \ b & = -v, \ c = x \\[6pt] & = w^3 - v^3 + x^3 + 3wvx \\[6pt] &= w^3 + (-v)^3 + x^3 - 3(w)(-v)(x) \\[6pt] &= [w + (-v) + x][w^2 + (-v)^2 + x^2 - (w)(-v) - (-v)(x) - (x)(w)] \\[6pt] &= (w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx) \\\\[12pt] & \color{magenta}\textbf{Denominator} \\[6pt] \color{magenta} (a + b + c)^2 &= \color{magenta} a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \\[6pt] &= w^2 + (-v)^2 + x^2 + 2(w)(-v) + 2(-v)(x) + 2(x)(w) \\[6pt] &= (w - v + x)^2 \\[6pt] & = (w - v + x) (w - v + x) \\ \\ \implies & \color{brown} \frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} \\[10pt] &= \frac{\cancel{(w - v + x)}(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x){\cancel{(w - v + x)}}} \\[10pt] &= \color{red}\mathbf{\frac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}} \end{aligned} \]

(iv) \( \dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] \color{magenta}a^2 - 2ab + b^2 &= \color{magenta} (a - b)^2 \\[6pt] & = 4y^2 - 20yz + 25z^2 \\[6pt] &= (2y)^2 - 2(2y)(5z) + (5z)^2 \\[6pt] &= (2y - 5z)^2 \\[6pt] &= (5z - 2y)(5z - 2y) \\\\[10pt] & \color{magenta}\textbf{Denominator} \\[6pt] \color{magenta} a^2 - b^2 & = \color{magenta} (a - b)(a + b) \\[6pt] & = 25z^2 - 4y^2 \\[6pt] &= (5z)^2 - (2y)^2 \\[6pt] &= (5z - 2y)(5z + 2y) \\\\[10pt] \implies & \color{brown} \frac{(5z - 2y)^2}{(5z - 2y)(5z + 2y)} \\[8pt] &= \frac{\cancel{(5z - 2y)}(5z - 2y)}{\cancel{(5z - 2y)}(5z + 2y)} \\[8pt] &= \color{red}\mathbf{\frac{5z - 2y}{5z + 2y}} \end{aligned} \]

(v) \( \dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] \text{Product } & \implies { \color{green}[3 \times (-2)]} = -6 \\ \text{Sum } & \implies {\color{green}(3-2)} = 1 \\ \text{Numbers } & \implies {\color{green}(3,-2)} \\[6pt] \color{brown} x^2 + x - 6 &= (x + 3)(x - 2) \\ \\[6pt] \text{Product } & \implies { \color{green}[(-3) \times (-4)]} = 12 \\ \text{Sum } & \implies {\color{green}(-3-4)} = -7 \\ \text{Numbers } & \implies {\color{green}(-3,-4)} \\[6pt] \color{brown} x^2 - 7x + 12 &= (x - 3)(x - 4) \\ \\[6pt] & \color{magenta}\textbf{Denominator} \\[6pt] \text{Product } & \implies { \color{green}[(-2) \times (-4)]} = 8 \\ \text{Sum } & \implies {\color{green}(-2-4)} = -6 \\ \text{Numbers } & \implies {\color{green}(-2,-4)} \\[6pt] \color{brown}x^2 - 6x + 8 &= (x - 2)(x - 4) \\\\[6pt] \color{brown} x^2 - 9 &= x^2 - 3^2 \\[6pt] & = (x - 3)(x + 3) \\\\[10pt] \implies & \color{brown} \frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} \\[10pt] &= \frac{\cancel{(x + 3)}\cancel{(x - 2)}\cancel{(x - 3)}\cancel{(x - 4)}}{\cancel{(x - 2)}\cancel{(x - 4)}\cancel{(x - 3)}\cancel{(x + 3)}} \\[8pt] &= \color{red}\mathbf{1} \end{aligned} \]

(vi) \( \dfrac{p^4 - 16}{p^2 - 4p + 4} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] p^4 - 16 &= (p^2)^2 - 4^2 \\[6pt] &= (p^2 - 4)(p^2 + 4) \\[6pt] &= (p^2 - 2^2)(p^2 + 4) \\[6pt] &= (p - 2)(p + 2)(p^2 + 4) \\\\[10pt] & \color{magenta}\textbf{Denominator} \\[6pt] p^2 - 4p + 4 &= (p - 2)^2 \\[6pt] & = (p - 2)(p - 2) \\\\[10pt] \implies & \color{brown} \frac{p^4 - 16}{p^2 - 4p + 4} \\[8pt] &= \frac{\cancel{(p - 2)}(p + 2)(p^2 + 4)}{(p - 2) {\cancel{(p - 2)}}} \\[8pt] &= \color{red}\mathbf{\frac{(p + 2)(p^2 + 4)}{p - 2}} \end{aligned} \]

Solution

(i) \( (-3x + 4)^2 \)

\[ \begin{aligned} \color{magenta} (a - b)^2 &= \color{magenta} a^2 - 2ab + b^2 \\[6pt] (-3x + 4)^2 &= (4 - 3x)^2 \\[6pt] &= (4)^2 - 2(4)(3x) + (3x)^2 \\[6pt] &= 16 - 24x + 9x^2 \\[6pt] &= \color{red}\mathbf{9x^2 - 24x + 16} \end{aligned} \]

(ii) \( (2s + 7)(2s - 7) \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) &= \color{magenta} a^2 - b^2 \\[6pt] (2s + 7)(2s - 7) &= (2s)^2 - (7)^2 \\[6pt] &= \color{red}\mathbf{4s^2 - 49} \end{aligned} \]

(iii) \( \left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right) \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) &= \color{magenta} a^2 - b^2 \\[6pt] \left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right) &= (p^2)^2 - \left(\frac{1}{2}\right)^2 \\[8pt] &= \color{red}\mathbf{p^4 - \frac{1}{4}} \end{aligned} \]

(iv) \( (2n + 7)(2n - 7) \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) &= \color{magenta} a^2 - b^2 \\[6pt] (2n + 7)(2n - 7) &= (2n)^2 - (7)^2 \\[6pt] &= \color{red}\mathbf{4n^2 - 49} \end{aligned} \]

(v) \( (s - 2t)(s^2 + 2st + 4t^2) \)

\[ \begin{aligned} \color{magenta} (a - b)(a^2 + ab + b^2) &= \color{magenta} a^3 - b^3 \\[6pt] (s - 2t)(s^2 + 2st + 4t^2) &= (s - 2t)\left[s^2 + (s)(2t) + (2t)^2\right] \\[6pt] &= (s)^3 - (2t)^3 \\[6pt] &= \color{red}\mathbf{s^3 - 8t^3} \end{aligned} \]

(vi) \( \left(\dfrac{1}{2r} - 4r\right)^2 \)

\[ \begin{aligned} \color{magenta} (a - b)^2 &= \color{magenta} a^2 - 2ab + b^2 \\[6pt] \left(\frac{1}{2r} - 4r\right)^2 &= \left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2 \\[8pt] &= \color{red}\mathbf{\frac{1}{4r^2} - 4 + 16r^2} \end{aligned} \]

(vii) \( (-3m + 4k - l)^2 \)

\[ \begin{aligned} \color{magenta} (a + b + c)^2 &= \color{magenta} a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \\[6pt] a = -3m, \ b &= 4k, \ c = -l \\[6pt] (-3m + 4k - l)^2 &= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m) \\[6pt] &= \color{red}\mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm} \end{aligned} \]

(viii) \( \left(x - \dfrac{1}{3}y\right)^3 \)

\[ \begin{aligned} \color{magenta} (a - b)^3 &= \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] \left(x - \frac{1}{3}y\right)^3 &= (x)^3 - 3(x)^2\left(\frac{1}{3}y\right) + 3(x)\left(\frac{1}{3}y\right)^2 - \left(\frac{1}{3}y\right)^3 \\[8pt] &= x^3 - \cancel{3}x^2\left(\frac{1}{\cancel{3}}y\right) + \cancel{3}x\left(\frac{1}{\cancel{9}_3}y^2\right) - \frac{1}{27}y^3 \\[8pt] &= \color{red} {x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3} \end{aligned} \]

(ix) \( \left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3 \)

\[ \begin{aligned} \color{magenta} (a - b)^3 &= \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] \left(\frac{7}{2}k - \frac{2}{3}m\right)^3 &= \left(\frac{7}{2}k\right)^3 - 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 - \left(\frac{2}{3}m\right)^3 \\[8pt] &= \frac{343}{8}k^3 - \cancel{3}\left(\frac{49}{\cancel{4}_2}k^2\right)\left(\frac{\cancel{2}}{\cancel{3}}m\right) + \cancel{3}\left(\frac{7}{\cancel{2}}k\right)\left(\frac{\cancel4^2}{\cancel9_3}m^2\right) - \frac{8}{27}m^3 \\[8pt] &= \color{red}\mathbf{\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3} \end{aligned} \]

Solution

(i) \( 17 \times 21 \)

\[ \begin{aligned} \color{magenta} (a - b)(a + b) &= \color{magenta} a^2 - b^2 \\[6pt] 17 \times 21 &= (19 - 2)(19 + 2) \\[6pt] &= (19)^2 - (2)^2 \\[6pt] &= 361 - 4 \\[6pt] &= \color{red}\mathbf{357} \end{aligned} \]

(ii) \( 104 \times 96 \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) &= \color{magenta} a^2 - b^2 \\[6pt] 104 \times 96 &= (100 + 4)(100 - 4) \\[6pt] &= (100)^2 - (4)^2 \\[6pt] &= 10000 - 16 \\[6pt] &= \color{red}\mathbf{9984} \end{aligned} \]

(iii) \( 24 \times 16 \)

\[ \begin{aligned} \color{magenta} (a + b)(a - b) &= \color{magenta} a^2 - b^2 \\[6pt] 24 \times 16 &= (20 + 4)(20 - 4) \\[6pt] &= (20)^2 - (4)^2 \\[6pt] &= 400 - 16 \\[6pt] &= \color{red}\mathbf{384} \end{aligned} \]

(iv) \( 147^3 \)

\[ \begin{aligned} \color{magenta} (a - b)^3 &= \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] 147^3 &= (150 - 3)^3 \\[6pt] &= (150)^3 - 3(150)^2(3) + 3(150)(3)^2 - (3)^3 \\[6pt] &= 3375000 - 9(22500) + 450(9) - 27 \\[6pt] &= 3375000 - 202500 + 4050 - 27 \\[6pt] &= \color{red}\mathbf{3176523} \end{aligned} \]

(v) \( 199^3 \)

\[ \begin{aligned} \color{magenta} (a - b)^3 &= \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] 199^3 &= (200 - 1)^3 \\[6pt] &= (200)^3 - 3(200)^2(1) + 3(200)(1)^2 - (1)^3 \\[6pt] &= 8000000 - 3(40000) + 600 - 1 \\[6pt] &= 8000000 - 120000 + 600 - 1 \\[6pt] &= \color{red}\mathbf{7880599} \end{aligned} \]

(vi) \( 127^3 \)

\[ \begin{aligned} \color{magenta} (a - b)^3 &= \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] 127^3 &= (130 - 3)^3 \\[6pt] &= (130)^3 - 3(130)^2(3) + 3(130)(3)^2 - (3)^3 \\[6pt] &= 2197000 - 9(16900) + 390(9) - 27 \\[6pt] &= 2197000 - 152100 + 3510 - 27 \\[6pt] &= \color{red}\mathbf{2048383} \end{aligned} \]

(vii) \( (-107)^3 \)

\[ \begin{aligned} (-107)^3 &= -(107)^3 \\[6pt] & = -(100 + 7)^3 \\[6pt] \color{magenta} -(a + b)^3 &= \color{magenta} -(a^3 + 3a^2b + 3ab^2 + b^3) \\[6pt] &= -\left[(100)^3 + 3(100)^2(7) + 3(100)(7)^2 + (7)^3\right] \\[6pt] &= -\left[1000000 + 21(10000) + 300(49) + 343\right] \\[6pt] &= -\left[1000000 + 210000 + 14700 + 343\right] \\[6pt] &= -(1225043) \\[6pt] & = \color{red}\mathbf{-1225043} \end{aligned} \]

(viii) \( (-299)^3 \)

\[ \begin{aligned} (-299)^3 &= -(299)^3 \\[6pt] & = -(300 - 1)^3 \\[6pt] &= -\left[(300)^3 - 3(300)^2(1) + 3(300)(1)^2 - (1)^3\right] \\[6pt] &= -\left[27000000 - 3(90000) + 900 - 1\right] \\[6pt] &= -\left[27000000 - 270000 + 900 - 1\right] \\[6pt] &= -(26730899) \\[6pt] & = \color{red}\mathbf{-26730899} \end{aligned} \]

Solution

(i) \( 4y^2 + 1 + \dfrac{1}{16y^2} \)

\[ \begin{aligned} \color{magenta} a^2 + 2ab + b^2 &= \color{magenta} (a + b)^2 \\[6pt] & = 4y^2 + 1 + \dfrac{1}{16y^2} \\[8pt] & = (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 \\[8pt] &= \color{red}\mathbf{\left(2y + \frac{1}{4y}\right)^2} \\[8pt] &= \color{green}\mathbf{\left(2y + \frac{1}{4y}\right)\left(2y + \frac{1}{4y}\right)} \end{aligned} \]

(ii) \( 9m^2 - \dfrac{1}{25n^2} \)

\[ \begin{aligned} \color{magenta} a^2 - b^2 &= \color{magenta} (a - b)(a + b) \\[6pt] & = 9m^2 - \dfrac{1}{25n^2} \\[8pt] & = (3m)^2 - \left(\frac{1}{5n}\right)^2 \\[8pt] &= \color{red}\mathbf{\left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)} \end{aligned} \]

(iii) \( 27b^3 - \dfrac{1}{64b^3} \)

\[ \begin{aligned} \color{magenta} a^3 - b^3 &= \color{magenta} (a - b)(a^2 + ab + b^2) \\[6pt] & = 27b^3 - \dfrac{1}{64b^3} \\[8pt] & = (3b)^3 - \left(\frac{1}{4b}\right)^3 \\[8pt] &= \left(3b - \frac{1}{4b}\right)\left[(3b)^2 + (3b)\left(\frac{1}{4b}\right) + \left(\frac{1}{4b}\right)^2\right] \\[8pt] &= \color{red}\mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)} \end{aligned} \]

(iv) \( x^2 + \dfrac{5x}{6} + \dfrac{1}{6} \)

\[ \begin{aligned} \text{Product } & \implies {\color{green}\frac{1}{2} \times \frac{1}{3}} = \frac{1}{6} \\[6pt] \text{Sum } & \implies {\color{green}\frac{1}{2} + \frac{1}{3}} = \frac{3+2}{6} = \frac{5}{6} \\[8pt] & = x^2 + \dfrac{5x}{6} + \dfrac{1}{6} \\[8pt] & = x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} \\[8pt] &= x\left(x + \frac{1}{2}\right) + \frac{1}{3}\left(x + \frac{1}{2}\right) \\[8pt] &= \color{red}{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)} \end{aligned} \]

(v) \( 27u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25} \)

\[ \begin{aligned} \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 &= \color{magenta} (a - b)^3 \\[6pt] 27u^3 &= (3u)^3 \\[6pt] \frac{1}{125} &= \left(\frac{1}{5}\right)^3 \\[8pt] \frac{27u^2}{5} &= 3(3u)^2\left(\frac{1}{5}\right) \\[8pt] \frac{9u}{25} &= 3(3u)\left(\frac{1}{5}\right)^2 \\\\[8pt] \implies & 27u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25} \\[8pt] &= (3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 \\[8pt] &= \color{red}\mathbf{\left(3u - \frac{1}{5}\right)^3} \end{aligned} \]

(vi) \( 64y^3 + \dfrac{1}{125}z^3 \)

\[ \begin{aligned} \color{magenta} a^3 + b^3 &= \color{magenta} (a + b)(a^2 - ab + b^2) \\[6pt] & = 64y^3 + \dfrac{1}{125}z^3 \\[8pt] & = (4y)^3 + \left(\frac{1}{5}z\right)^3 \\[8pt] &= \left(4y + \frac{1}{5}z\right)\left[(4y)^2 - (4y)\left(\frac{1}{5}z\right) + \left(\frac{1}{5}z\right)^2\right] \\[8pt] &= \color{red}\mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)} \end{aligned} \]

(vii) \( p^3 + 27q^3 + r^3 - 9pqr \)

\[ \begin{aligned} \color{magenta} a^3 + b^3 + c^3 - 3abc &= \color{magenta} (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\[6pt] p^3 + 27q^3 + r^3 - 9pqr & = (p)^3 + (3q)^3 + (r)^3 - 3(p)(3q)(r)\\[6pt] a &= p \ , b = 3q \ , \ c = r \\[6pt] &= \color{red}\mathbf{(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)} \end{aligned} \]

(viii) \( 9m^2 - 12m + 4 \)

\[ \begin{aligned} \color{magenta} a^2 - 2ab + b^2 &= \color{magenta} (a - b)^2 \\[6pt] & = 9m^2 - 12m + 4 \\[6pt] & = (3m)^2 - 2(3m)(2) + (2)^2 \\[6pt] &= \color{red}\mathbf{(3m - 2)^2} \\[6pt] &= \color{green}\mathbf{(3m - 2)(3m - 2)} \end{aligned} \]

(ix) \( 9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz \)

\[ \begin{aligned} \color{magenta}\textbf{Taking out } & \frac{1}{3} \color{magenta} \textbf{ as a common factor} \\[6pt] &= \frac{1}{3}\left(27x^3 - 8y^3 + z^3 + 18xyz\right) \\[8pt] &= \frac{1}{3}\left[(3x)^3 + (-2y)^3 + (z)^3 - 3(3x)(-2y)(z)\right] \\\\[8pt] a &= 3x \ , b = (-2y) \ , \ c = z \\\\[6pt] \color{magenta} a^3 + b^3 + c^3 - 3abc &= \color{magenta} (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\[6pt] &= {\frac{1}{3}[3x +(- 2y) + z][(3x)^2 + (-2y)^2 + (z)^2 - (3x)(-2y) - (-2y)(z) - (3x)(z)]} \\[6pt] &= \color{red} {\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz)} \end{aligned} \]

(x) \( 4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz \)

\[ \begin{aligned} \color{magenta} (a + b + c)^2 & = \color{magenta} a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \\[6pt] a & = 2x, \ b = 3y, \ c = 6z \\[6pt] &= (2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x) \\[6pt] &= \color{red}{(2x + 3y + 6z)^2} \\[6pt] &= \color{green} {(2x + 3y + 6z)(2x + 3y + 6z)} \end{aligned} \]

(xi) \( 27u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4} \)

\[ \begin{aligned} \color{magenta} (a - b)^3 & = \color{magenta} a^3 - 3a^2b + 3ab^2 - b^3 \\[6pt] a & \implies 27u^3 = (3u)^3 \\[6pt] b & \implies \frac{1}{216} = \left(\frac{1}{6}\right)^3 \\[8pt] \implies & 27u^3 - \dfrac{9u^2}{2} + \dfrac{u}{4} - \dfrac{1}{216} \\[8pt] & = (3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3 \\[8pt] &= \color{red}\mathbf{\left(3u - \frac{1}{6}\right)^3} \end{aligned} \]

Solution

(i) \( \dfrac{4x^2 + 4x + 1}{4x^2 - 1} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] & = 4x^2 + 4x + 1 \\[6pt] & = (2x)^2 + 2(2x)(1) + (1)^2 \\[6pt] &= (2x + 1)^2 \\[6pt] & = (2x + 1)(2x + 1) \\\\[8pt] & \color{magenta}\textbf{Denominator} \\[6pt] & = 4x^2 - 1 \\[6pt] &= (2x)^2 - 1^2 \\[6pt] & = (2x - 1)(2x + 1) \\\\[8pt] \implies & \dfrac{4x^2 + 4x + 1}{4x^2 - 1} \\[6pt] & = \frac{(2x + 1)\cancel{(2x + 1)}}{(2x - 1)\cancel{(2x + 1)}} \\[6pt] & = \color{red} {\frac{2x + 1}{2x - 1}} \end{aligned} \]

(ii) \( \dfrac{9(3a^3 - 24b^3)}{9a^2 - 36b^2} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] 9(3a^3 - 24b^3) &= 9 \times 3(a^3 - 8b^3) \\[6pt] &= 27\left[a^3 - (2b)^3\right] \\[6pt] &= 27(a - 2b)(a^2 + 2ab + 4b^2) \\\\[8pt] & \color{magenta}\textbf{Denominator} \\[6pt] 9a^2 - 36b^2 &= 9(a^2 - 4b^2) = 9(a - 2b)(a + 2b) \\\\[8pt] \implies & \frac{\cancel{27}^3\cancel{(a - 2b)}(a^2 + 2ab + 4b^2)}{\cancel9_1\cancel{(a - 2b)}(a + 2b)} \\[10pt] &= \color{red}\mathbf{\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}} \end{aligned} \]

(iii) \( \dfrac{s^3 + 125t^3}{s^2 - 2st - 35t^2} \)

\[ \begin{aligned} & \color{magenta}\textbf{Numerator} \\[6pt] s^3 + 125t^3 &= s^3 + (5t)^3 = (s + 5t)(s^2 - 5st + 25t^2) \\\\[8pt] & \color{magenta}\textbf{Denominator} \\[6pt] s^2 - 2st - 35t^2 &= s^2 - 7st + 5st - 35t^2 \\[6pt] &= s(s - 7t) + 5t(s - 7t) \\[6pt] &= (s - 7t)(s + 5t) \\\\[8pt] \implies & \frac{\cancel{(s + 5t)}(s^2 - 5st + 25t^2)}{(s - 7t)\cancel{(s + 5t)}} = \color{red}\mathbf{\frac{s^2 - 5st + 25t^2}{s - 7t}} \end{aligned} \]

Solution

(i) \( 25a^2 - 30ab + 9b^2 \)

\[ \begin{aligned} \text{Area} &= 25a^2 - 30ab + 9b^2 \\[6pt] &= (5a)^2 - 2(5a)(3b) + (3b)^2 \\[6pt] &= (5a - 3b)^2 \\[6pt] &= (5a - 3b)(5a - 3b) \\\\[8pt] \color{red}\mathbf{\text{Length}} &= \color{red}\mathbf{5a - 3b} \\[6pt] \color{red}\mathbf{\text{Breadth}} &= \color{red}\mathbf{5a - 3b} \end{aligned} \]

(ii) \( 36s^2 - 49t^2 \)

\[ \begin{aligned} \text{Area} &= 36s^2 - 49t^2 \\[6pt] &= (6s)^2 - (7t)^2 \\[6pt] &= (6s + 7t)(6s - 7t) \\\\[8pt] \color{red}\mathbf{\text{Length}} &= \color{red}\mathbf{6s + 7t} \\[6pt] \color{red}\mathbf{\text{Breadth}} &= \color{red}\mathbf{6s - 7t} \end{aligned} \]

Solution

(i) \( 6a^2 - 24b^2 \)

\[ \begin{aligned} \text{Volume} &= 6(a^2 - 4b^2) \\[6pt] &= 6\left[a^2 - (2b)^2\right] \\[6pt] &= 6(a - 2b)(a + 2b) \\\\[8pt] \color{red}\mathbf{\text{Dimensions: }} & \color{red}\mathbf{6, \quad (a - 2b), \quad (a + 2b)} \end{aligned} \]

(ii) \( 3ps^2 - 15ps + 12p \)

\[ \begin{aligned} \text{Volume} &= 3p(s^2 - 5s + 4) \\[6pt] &= 3p(s^2 - 4s - s + 4) \\[6pt] &= 3p[s(s - 4) - 1(s - 4)] \\[6pt] &= 3p(s - 1)(s - 4) \\\\[8pt] \color{red}\mathbf{\text{Dimensions: }} & \color{red}\mathbf{3p, \quad (s - 1), \quad (s - 4)} \end{aligned} \]

Solution

\[ \begin{aligned} \text{Side of the inner square playground} &= 40\text{ m} \\[6pt] \text{Area of playground} &= 40 \times 40 \\[6pt] & = 1600\text{ m}^2 \\\\[8pt] \text{Side of outer square including path} &= 40 + 2s\text{ m} \\[6pt] \text{Total outer area} &= (40 + 2s)^2 \\[6pt] &= (40)^2 + 2(40)(2s) + (2s)^2 \\[6pt] &= 1600 + 160s + 4s^2 \\\\[10pt] \text{Area of the path} &= \text{Outer Area} - \text{Inner Area} \\[6pt] &= (1600 + 160s + 4s^2) - 1600 \\[6pt] &= \color{red}\mathbf{4s^2 + 160s\text{ m}^2} \\[6pt] &= \color{green}\mathbf{4s(s + 40)\text{ m}^2} \end{aligned} \]

Answer The area of the path is \( \color{red}\mathbf{(4s^2 + 160s)\text{ m}^2} \).

Solution

\[ \begin{aligned} \text{Let the number} & \text{ be } x \\[6pt] x + \frac{1}{x} &= \frac{10}{3} \\[8pt] \frac{x^2 + 1}{x} &= \frac{10}{3} \\[8pt] 3(x^2 + 1) &= 10x \\[6pt] 3x^2 + 3 &= 10x \\[6pt] 3x^2 - 10x + 3 &= 0 \\[8pt] 3x^2 - 9x - x + 3 &= 0 \\[6pt] 3x(x - 3) - 1(x - 3) &= 0 \\[6pt] (3x - 1)(x - 3) &= 0 \\\\[8pt] \implies x - 3 = 0 \ & \text{ (or) } 3x - 1 = 0 \\[6pt] \implies \color{red} {x = 3} \quad &\text{(or)} \quad \color{red} {x = \frac{1}{3}} \end{aligned} \]

Answer The number is \( \color{red}\mathbf{3} \) and \( \color{red}\mathbf{\dfrac{1}{3}} \).

Solution

\[ \begin{aligned} \text{Area} &= 2x^2 + 7x + 3 \\[6pt] \text{Width} &= 2x + 1 \\\\[8pt] \text{Area} & = 2x^2 + 7x + 3 \\[6pt] &= 2x^2 + 6x + x + 3 \\[6pt] &= 2x(x + 3) + 1(x + 3) \\[6pt] \text{Area} &= (2x + 1)(x + 3) \\[10pt] \text{Length} &= \frac{\text{Area}}{\text{Width}} \\[8pt] &= \frac{\cancel{(2x + 1)}(x + 3)}{\cancel{2x + 1}} \\[8pt] \color{red} \text{Length} &= \color{red} {(x + 3)\text{ hastas}} \end{aligned} \]

Answer The length of the pool is \( \color{red}\mathbf{(x + 3)\text{ hastas}} \).

Solution

\[ \begin{aligned} \text{Let } P(x) &= px^2 + 5x + r \\\\[8pt] \color{magenta}\textbf{Since } (x - 2)& \color{magenta} \textbf{ is a factor: } \\[8pt] P(2) &= 0 \\[6pt] p(2)^2 + 5(2) + r &= 0 \\[6pt] 4p + 10 + r &= 0 \\[6pt] 4p + r &= -10 \ \color{magenta} \longrightarrow \enclose{circle}{1} \\\\[10pt] \color{magenta}\textbf{Since } \left(x - \frac{1}{2}\right) & \color{magenta} \textbf{ is a factor: } \\[8pt] P\left(\frac{1}{2}\right) &= 0 \\[6pt] p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r &= 0 \\[8pt] \frac{p}{4} + \frac{5}{2} + r &= 0 \\[8pt] p + 10 + 4r &= 0 \\[6pt] p + 4r &= -10 \ \color{magenta} \longrightarrow \enclose{circle}{2} \\\\[10pt] \color{magenta}\textbf{Equating } \enclose{circle}{1} \textbf{ and } \enclose{circle}{2}\textbf{:} & \\[6pt] 4p + r &= p + 4r \\[6pt] 4p - p &= 4r - r \\[6pt] 3p &= 3r \\[6pt] \color{red}\mathbf{p} &= \color{red}\mathbf{r} \\[10pt] \textbf{Hence proved} \end{aligned} \]

Answer Proved that \( \color{red}\mathbf{p = r} \).

Solution

\[ \begin{aligned} \color{magenta}\textbf{Given: } \\ a + b + c &= 5 \\ ab + bc + ca &= 10 \\\\[8pt] \color{magenta} (a + b + c)^2 &= \color{magenta} a^2 + b^2 + c^2 + 2(ab + bc + ca) \\[6pt] (5)^2 &= a^2 + b^2 + c^2 + 2(10) \\[6pt] 25 &= a^2 + b^2 + c^2 + 20 \\[6pt] a^2 + b^2 + c^2 &= 25 - 20 \\[6pt] a^2 + b^2 + c^2 &= \color{blue}\mathbf{5} \\\\[12pt] \color{magenta}\textbf{Identity:} & \\[6pt] \color{magenta}a^3 + b^3 + c^3 - 3abc &= \color{magenta} (a + b + c)\left[(a^2 + b^2 + c^2) - (ab + bc + ca)\right] \\[8pt] &= 5 \times [5 - 10] \\[6pt] &= 5 \times (-5) \\[6pt] &= \color{red}\mathbf{-25} \\[10pt] & \textbf{Hence proved.} \end{aligned} \]

Answer Proved that \( \color{red}\mathbf{a^3 + b^3 + c^3 - 3abc = -25} \).

Solution

\[ \begin{aligned} n^3 - n &= n(n^2 - 1) \\[6pt] &= \color{blue}\mathbf{n(n - 1)(n + 1)} \\\\[8pt] \text{Notice that } (n - 1), \ n, \ (n + 1) & \text{ are three consecutive integers.} \\\\[6pt] \textbf{1. Divisibility by 2: } & \text{Out of any two consecutive integers, at least} \\ & \text{one is even. Thus, the product is divisible by 2.} \\[6pt] \textbf{2. Divisibility by 3: } & \text{Out of any three consecutive integers, exactly} \\ & \text{one is a multiple of 3. Thus, the product is divisible by 3.} \\\\[8pt] \implies \text{Since product of } n(n - 1)&(n + 1) \text{ is always } \text{ divisible by } 2 \times 3 = \mathbf{6}. \\\\[6pt] \color{red} n^3 - n & \color{red}\textbf{ is always divisible by } 6 \end{aligned} \]

Answer Factoring gives \( \color{red}\mathbf{(n - 1)n(n + 1)} \), which is the product of three consecutive integers and is therefore always divisible by both 2 and 3, hence divisible by \( \color{red}\mathbf{6} \).

Solution

(i) \( x^3 + y^3 - 12xy + 64 \) , when \( x + y = -4 \)

\[ \begin{aligned} x + y &= -4 \\[8pt] \text{Cubing on } & \text{both sides} \\[8pt] (x + y)^3 &= (-4)^3 \\[8pt] x^3 + y^3 + 3xy(x+y) &= -64 \\[8pt] x^3 + y^3 + 3xy(-4) &= -64 \\[8pt] x^3 + y^3 - 12xy &= -64 \\[8pt] \implies \color{red} x^3 + y^3 - 12xy +64 &= \color{red}\mathbf{0} \end{aligned} \]

(ii) \( x^3 - 8y^3 - 36xy - 216 \) , when \( x = 2y + 6 \)

\[ \begin{aligned} x &= 2y + 6 \\[8pt] x -2y &= 6 \\[8pt] \text{Cubing on } & \text{both sides} \\[8pt] (x - 2y)^3 &= (6)^3 \\[8pt] x^3 - 8y^3 - 3x(2y)(x - 2y) &= 216 \\[8pt] x^3 - 8y^3 - 6xy(6) &= 216 \\[8pt] x^3 - 8y^3 - 36xy &= 216 \\[8pt] \implies \color{red} x^3 - 8y^3 - 36xy - 216 &= \color{red}\mathbf{0} \end{aligned} \]