Solution

Answer \( {\color{orange}(iii)}\ \color{red}{(3z-1)^2} \)

Answer \( {\color{orange}(iv)}\ \color{red}{(a-b)^2=a^2-2ab+b^2} \)

Solution

Answer \( {\color{orange}(iii)}\ \color{red}{(y+1)(y+3)} \)

Solution

Answer \( {\color{orange}(i)}\ \color{red}{(p-19)(p+2)} \)

Solution

Answer \( {\color{orange}(iii)}\ \color{red}{(2x+3)\ \text{units}} \)

Solution

\[ \begin{aligned} & \big[ {\color{magenta}{(a+b)(a-b)=a^2-b^2}}\big] \\[8pt] &= (a+3)(a-3)(a^2+9) \\[4pt] &= \left[(a)^2-(3)^2 \right](a^2+9) \\[4pt] &= (a^2-9) (a^2+9) \\[4pt] &= (a^2)^2-(9)^2 \\[4pt] &= a^4-81 \end{aligned} \]

Answer \(=\ \color{red}{a^4-81}\)

Solution

\[ \begin{aligned} \color{magenta} (a+b)^2 &= \color{magenta}a^2+b^2 +2ab \\[8pt] \left(x^2+\frac{1}{x^2}\right)^2 &= x^4+2+\frac{1}{x^4} \\[4pt] 6^2 &= x^4+2+\frac{1}{x^4} \\[4pt] 36 &= x^4+2+\frac{1}{x^4} \\[4pt] 36 -2 &= x^4 + \frac{1}{x^4} \\[4pt] 34 &= x^4 + \frac{1}{x^4} \\[4pt] \end{aligned} \]

Answer \( x^4+\dfrac{1}{x^4}=\ \color{red}{34}\)

Solution

\[ \begin{aligned} & \big[ {\color{magenta}{a^2-b^2=(a-b)(a+b)}}\big] \\[8pt] xy^2k & = (4xy+3y)^2-(4xy-3y)^2 \\[4pt] xy^2k &= \big[4xy+3y-( 4xy - 3y) \big] \big[4xy+3y + (4xy-3y )\big] \\[4pt] xy^2k &= \big(\cancel{4xy} +3y - \cancel{4xy} + 3y \big) \big(4xy + \cancel{3y} + 4xy- \cancel{3y}\big) \\[4pt] xy^2k &= (6y)(8xy) \\[4pt] xy^2k &= 48xy^2 \\[6pt] k &= \frac{48 \cancel{xy^2}}{\cancel{xy^2}} \\[6pt] k &= 48 \\[4pt] \end{aligned} \]

Answer \(k=\ \color{red}{48}\)

Solution

\[ \begin{aligned} (a+b)^2 &= a^2 + b^2 + 2ab \\ &= 9+ 2(4) \\ &= 9+8 \\ \color{magenta} (a + b)^2 & = 17 \\[5pt] (a-b)^2 &= a^2 + b^2 -2ab \\ &= 9- 2(4) \\ &= 9-8 \\ \color{magenta} (a - b)^2 &= 1 \\[5pt] 3(a+b)^2-2(a-b)^2 &= 3 (17) - 2 (1) \\ &= 51-2 \\ 3(a+b)^2-2(a-b)^2 &= 49 \end{aligned} \]

Answer \(3(a+b)^2-2(a-b)^2=\ \color{red}{49}\)

Solution

\[ \begin{aligned} (a+b+c)^2 &= a^2+b^2+c^2+2(ab+bc+ac) \\[4pt] (12)^2 &= 64+2(ab+bc+ac) \\[4pt] 144 &= 64+2(ab+bc+ac) \\[4pt] 144 - 64 &= 2(ab+bc+ac) \\[4pt] 80 &= 2(ab+bc+ac) \\[6pt] \frac{80}{2} &= ab+bc+ac \\[6pt] 40 &= ab+bc+ac \\[6pt] \end{aligned} \]

Answer \(ab+bc+ac=\ \color{red}{40}\)

Solution

\[ \begin{aligned} & \big[ {\color{magenta}{(a-b)^2 = a^2 + b^2 - 2ab}} \big] \\[8pt] (7x - 9y)(7x - 9y) &= (7x - 9y)^2 \\[4pt] &= (7x)^2 + (9y)^2 - 2 (7x)(9y) \\[4pt] &= 49x^2 + 81y^2 - 126xy \end{aligned} \]

Answer \( \ \color{red} {49x^2 + 81y^2 - 126xy} \)

Solution

\[ \begin{aligned} \color{magenta}(a+b)(a-b) & = \color{magenta} a^2-b^2 \\[6pt] (6ax + 5by)(6ax - 5by) &= (6ax)^2 - (5by)^2 \\[4pt] &= 36a^2x^2 - 25b^2y^2 \end{aligned} \]

Answer \( \color{red}{36a^2x^2 - 25b^2y^2} \)

Solution

\[ \begin{aligned} \color{magenta} (a+b)^2 &= \color{magenta} a^2+b^2+2ab \\[6pt] & = \left(\frac{3}{4}p^2+\frac{2}{3}q^2\right)^2 \\[6pt] &= \left(\frac{3}{4}p^2\right)^2 + \left(\frac{2}{3}q^2\right)^2 + 2 \left(\frac{3}{4}p^2\right) \left(\frac{2}{3}q^2\right) \\[6pt] &= \frac{9}{16}p^4 + \frac{4}{9}q^4 + p^2q^2 \end{aligned} \]

Answer \( \color{red}{\dfrac{9}{16}p^4 + p^2q^2 + \dfrac{4}{9}q^4} \)

Solution

\[ \begin{aligned} & \color{magenta}{(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc} \\[6pt] &= x^2+\left(\frac{3}{4}y\right)^2+(-4z)^2 +2(x)\!\left(\frac{3}{4}y\right)+2(x)(-4z)+2\!\left(\frac{3}{4}y\right)(-4z) \\[6pt] &= x^2+\frac{9}{16}y^2+16z^2+\frac{3}{2}xy-8xz-6yz \end{aligned} \]

Answer \( \color{red}{x^2+\dfrac{9}{16}y^2+16z^2+\dfrac{3}{2}xy-8xz-6yz} \)

Solution

\[ \begin{aligned} & \color{magenta}(a+b)(a-b) = a^2-b^2 \\[6pt] & = (x+2y)(x-2y)(x^2+4y^2) \\[4pt] &= \left[(x)^2-(2y)^2 \right] (x^2+4y^2) \\[4pt] &= (x^2 -4y^2) (x^2+4y^2) \\[4pt] &= (x^2)^2 -(4y^2)^2 \\[4pt] &= x^4-16y^4 \end{aligned} \]

Answer \( \color{red}{x^4-16y^4} \)

Solution

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

Answer \( \color{red}{x^8-\dfrac{1}{x^8}} \)

Solution

\[ \begin{aligned} & \color{magenta}{(a+b)^2=a^2+b^2+2ab} \\[6pt] & = 4x^2+12xy+9y^2 \\[6pt] &= (2x)^2 + 2(2x)(3y) + (3y)^2 \\[6pt] &= \color{green}{(2x+3y)^2} \\ \\ & x=-2,\ y=2 \\ \\ & = (2x+3y)^2 \\[6pt] &= \big[2(-2)+3(2)\big]^2 \\[6pt] &= (-4+6)^2 \\[6pt] & = (2)^2 \\[6pt] &=4 \end{aligned} \]

Answer \( \color{red}{(2x+3y)^2} \ ; \ 4\)

Solution

\[ \begin{aligned} & \color{magenta}{(a-b)^2=a^2+b^2-2ab} \\[6pt] & = \frac{4}{25}m^2 - mn + \frac{25}{16}n^2 \\[6pt] &= \left(\frac{2}{5}m\right)^2 - 2\left(\frac{2}{5}m\right)\!\left(\frac{5}{4}n\right) + \left(\frac{5}{4}n\right)^2\\[6pt] &= \color{green}{\left(\frac{2}{5}m - \frac{5}{4}n\right)^2} \\ \\ & m=5,\ n=8 \\ \\ &= \left[\frac{2}{\cancel5_1}(\cancel5^1) - \frac{5}{\cancel4_1} (\cancel8^2)\right]^2 \\[6pt] &= (2 - 10)^2 \\[6pt] &= (-8)^2 \\[6pt] &= 64 \end{aligned} \]

Answer \( \color{red}{\left(\dfrac{2}{5}m - \dfrac{5}{4}n\right)^2} \ ; \ 64\)

Solution

\[ \begin{aligned} &=(a+b)^2+(a-b)^2 \\[4pt] &= (a^2+b^2+2ab)+(a^2+b^2-2ab) \\[4pt] &= a^2+b^2+ \cancel{2ab} + a^2+b^2-\cancel{2ab} \\[4pt] &= 2a^2+2b^2 \\[2pt] &= 2(a^2+b^2) \end{aligned} \]

Answer \( \color{red}{2(a^2+b^2)} \)

Solution

\[ \begin{aligned} &=(3x+7y)^2-(3x-7y)^2 \\[4pt] &=\big[(3x)^2+(7y)^2+2(3x)(7y)\big] \\[4pt] &=\big[9x^2+49y^2+ 42xy \big] -\big[9x^2+49y^2-42xy\big] \\[4pt] &=\cancel{9x^2}+\cancel{49y^2}+42xy-\cancel{9x^2}-\cancel{49y^2}+42xy \\[4pt] &=84xy \end{aligned} \]

Answer \( \color{red}{84xy} \)

Solution

\[ \begin{aligned} &=(6a+5b+4c)^2-(6a-5b-4c)^2 \\[4pt] &=\big[36a^2+25b^2+16c^2+60ab+48ac+40bc\big] \\ &\quad-\big[36a^2+25b^2+16c^2-60ab-48ac+40bc\big] \\[4pt] &=\cancel{36a^2}+\cancel{25b^2}+\cancel{16c^2}+60ab+48ac+\cancel{40bc} \\ &\quad-\cancel{36a^2}-\cancel{25b^2}-\cancel{16c^2}+60ab+48ac-\cancel{40bc} \\[4pt] &=120ab+96ac \end{aligned} \]

Answer \( \color{red}{120ab+96ac} \)

Solution

\[ \begin{aligned} &=(a^2-b^2)(a^2+b^2)-(a^2-b^2)^2 \\[4pt] &=\big[(a^2)^2-(b^2)^2\big]-\big[(a^2)^2-2(a^2)(b^2)+(b^2)^2 \big] \\[4pt] &=\big(a^4-b^4 \big) - \big(a^4-2a^2b^2+b^4 \big) \\[4pt] &=\cancel{a^4}-b^4-\cancel{a^4}+2a^2b^2-b^4 \\[4pt] &= 2a^2b^2-2b^4 \end{aligned} \]

Answer \( \color{red} 2a^2b^2-2b^4 \)

Solution

\[ \begin{aligned} & \color{magenta}{(a-b)^2=a^2+b^2-2ab} \\[4pt] (399)^2&=(400-1)^2\\ &=400^2+1^2-2(400)(1)\\ &=160000+1-800\\ &=159201 \end{aligned} \]

Answer \(\color{red}{159201}\)

Solution

\[ \begin{aligned} & \color{magenta}{(a+b)^2=a^2+b^2+2ab} \\[4pt] (62)^2&=(60+2)^2\\ &=60^2+2^2+2(60)(2)\\ &=3600+4+240\\ &=3844 \end{aligned} \]

Answer \(\color{red}{3844}\)

Solution

\[ \begin{aligned} & \color{magenta}{(a+b)(a-b)=a^2-b^2} \\[4pt] 103\times97&=(100+3)(100-3)\\ &=100^2-3^2\\ &=10000-9\\ &=9991 \end{aligned} \]

Answer \(\color{red}{9991}\)

Solution

\[ \begin{aligned} & \color{magenta}{(a+b)^2=a^2+b^2+2ab} \\[4pt] (10.1)^2&=(10+0.1)^2\\ &=10^2+(0.1)^2+2(10)(0.1)\\ &=100+0.01+2\\ &=102.01 \end{aligned} \]

Answer \(\color{red}{102.01}\)

Solution

\[ \begin{aligned} \color{magenta} a^2-b^2 &= \color{magenta} (a-b)(a+b) \\[4pt] (85)^2-(75)^2&=(85-75)(85+75)\\ &=10\times160\\ &=1600 \end{aligned} \]

Answer \(\color{red}{1600}\)

Solution

\[ \begin{aligned} & \color{magenta}{(a-b)(a+b)=a^2-b^2} \\[4pt] 291\times309&=(300-9)(300+9)\\ &=300^2-9^2\\ &=90000-81\\ &=89919 \end{aligned} \]

Answer \(\color{red}{89919}\)

Solution

\[ \begin{aligned} & \color{magenta}{a^2-b^2=(a-b)(a+b)} \\[4pt] 36x &= 78^2-42^2 \\[6pt] 36x &= (78-42)(78+42) \\[6pt] 36x &= 36\times 120 \\[6pt] 36x &= 4320 \\[6pt] x &= \frac{4320}{36} \\[6pt] x &=\color{green}{120} \end{aligned} \]

Answer \( \color{red}{x=120} \)

Solution

\[ \begin{aligned} \color{magenta} a^2-b^2 &= \color{magenta} (a-b)(a+b) \\[6pt] 6.2x &= (8.1)^2-(1.9)^2 \\[6pt] 6.2x &= (8.1-1.9)(8.1+1.9) \\[6pt] 6.2x &= 6.2\times 10 \\[6pt] x &= \frac{\cancel{6.2} \times 10}{\cancel{6.2}}\\[6pt] x&= \color{green}{10} \end{aligned} \]

Answer \( \color{red}{x=10} \)

Solution

\[ \begin{align*} &= z^2\ {\color{magenta}-\ 4z}\ - 77 \\[4pt] &= z^2\ {\color{magenta}-\ 11z + 7z}\ - 77 \\[4pt] &= z(z-11) + 7(z-11) \\[4pt] &= \color{red}{(z-11)(z+7)} \end{align*} \]

Solution

\[ \begin{aligned} &= x^2\ {\color{magenta}+\,25x}\ + 144 \\[4pt] &= x^2\ {\color{magenta}+\,9x+16x}\ + 144 \\[4pt] &= x(x+9)+16(x+9) \\[4pt] &= \color{red}{(x+9)(x+16)} \end{aligned} \]

Solution

\[ \begin{aligned} &= x^2\ {\color{magenta}+\,5x}\ -104 \\[4pt] &= x^2\ {\color{magenta}+\,13x-8x}\ -104 \\[4pt] &= x(x+13)-8(x+13) \\[4pt] &= \color{red}{(x+13)(x-8)} \end{aligned} \]

Solution

\[ \begin{aligned} &= 49x^2 - 64 \\[2pt] &= (7x)^2 - (8)^2 \\[2pt] &= \color{red}{(7x-8)(7x+8)} \end{aligned} \]

Solution

\[ \begin{aligned} &= x^2 - 1 - 2y - y^2 \\[4pt] &= x^2 - \big(y^2+2y+1\big) \\[4pt] &= x^2 - \big[(y)^2+2(1)(y)+(1)^2 \big] \\[4pt] &= x^2 - (y+1)^2 \\[4pt] &= [x-(y+1)][x+(y+1)] \\[4pt] &= \color{red}{(x-y-1)(x+y+1)} \end{aligned} \]

Solution

\[ \begin{aligned} & = 4(x-y)^2-12(x-y)+9 \\[4pt] &= [2{x-y}]^2-2[2(x-y)](3)+(3)^2 \\[4pt] &= {\big[2(x-y)-3\big]^2}\\[4pt] &= \color{red} \big[2(x-y)-3\big]\big[2(x-y)-3\big] \end{aligned} \]

Solution

\[ \begin{aligned} &= x^2-x - xy + y \\[4pt] &= x(x-1)-y(x-1) \\[4pt] &= \color{red}{(x-1)(x-y)} \end{aligned} \]

Solution

\[ \begin{aligned} & = 81(x+1)^2 + 90(x+1)(y+2) + 25(y+2)^2 \\[4pt] & = [9(x+1)]^2 + 2 \times 9(x+1) \times 5(y+2) + [5(y+2)]^2 \\[8pt] &\color{magenta}{a^2+2ab+b^2=(a+b)^2} \\[4pt] a &=9(x+1),\ \ b=5(y+2) \\[2pt] &= \big[9(x+1)+5(y+2)\big]^2\\[4pt] &= \big[9x+ 9 +5y+10 \big]^2\\[4pt] &= \big[9x+5y+19 \big]^2\\[4pt] &= \color{red} \big(9x+5y+19 \big)\big(9x+5y+19 \big) \end{aligned} \]

Solution

\[ \begin{aligned} & = x^2 + y^2 + \dfrac{z^2}{4} + 2xy - yz - zx \\[8pt] & = (x)^2 + (y)^2 + \left(\frac{-z}{2}\right)^2 + 2(x)(y) + 2 (y) \left(\frac{-z}{2}\right) + 2 \left(\frac{-z}{2}\right) (x) \\[8pt] &\color{magenta}{(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc} \\[6pt] a&=x,\ b=y,\ c=\frac{-z}{2} \\[4pt] &= {\left[x+y-\frac{z}{2}\right]^2} \\[4pt] &= \color{red} \left[x+y-\frac{z}{2}\right] \left[x+y-\frac{z}{2}\right] \end{aligned} \]

Solution

\[ \begin{aligned} \left(x+\frac{1}{x}\right) &= 8 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(x+\frac{1}{x}\right)^2 &= 8^2 \\[6pt] x^2+\frac{1}{x^2}+2 \times \cancel{x^2} \times \frac{1}{\cancel{x^2}} &= 64 \\[6pt] x^2+\frac{1}{x^2} + 2 &= 64 \\[6pt] x^2+\frac{1}{x^2} &= 64 -2 \\[6pt] x^2+\frac{1}{x^2} &= 62 \\[6pt] \end{aligned} \]

Answer \( x^2+\dfrac{1}{x^2} = \color{red}{62} \)

Solution

\[ \begin{aligned} \left(a-\frac{1}{a}\right) &= 5 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= 5^2 \\[6pt] a^2+\frac{1}{a^2}-2 \times \cancel{a^2} \times \frac{1}{\cancel{a^2}} &= 25\\[6pt] a^2+\frac{1}{a^2}-2 &= 25\\[6pt] a^2+\frac{1}{a^2} &= 25 + 2\\[6pt] a^2+\frac{1}{a^2} &= 27 \end{aligned} \]

Answer \( a^2+\dfrac{1}{a^2} = \color{red}{27} \)

Solution

\[ \begin{aligned} \left(a+\frac{1}{a}\right)^2 &= a^2+\frac{1}{a^2}+2 \times \cancel{a^2} \times \frac{1}{\cancel{a^2}} \\[6pt] \left(a+\frac{1}{a}\right)^2 &= a^2+\frac{1}{a^2}+2 \\[6pt] \left(a+\frac{1}{a}\right)^2 &= 23+2 \\[6pt] \left(a+\frac{1}{a}\right)^2 &= 25 \\[6pt] a+\frac{1}{a} &= 5 \end{aligned} \]

Answer \( \left(a+\dfrac{1}{a}\right) = \color{red}{5}\)

Solution

\[ \begin{aligned} \left(x-\frac{1}{x}\right)^2 &= x^2+\frac{1}{x^2}-2 \times \cancel{x^2} \times \frac{1}{\cancel{x^2}} \\[6pt] \left(x-\frac{1}{x}\right)^2 &= x^2+\frac{1}{x^2}-2 \\[6pt] \left(x-\frac{1}{x}\right)^2 &= 51-2 \\[6pt] \left(x-\frac{1}{x}\right)^2 &= 49 \\[6pt] x-\frac{1}{x} &= 7 \end{aligned} \]

Answer \( \left(x-\dfrac{1}{x}\right) = \color{red}{7}\)

Solution

\[ \begin{aligned} & = \frac{2.3\times2.3-0.3\times0.3}{2.3\times2.3-2\times2.3\times0.3+0.3\times0.3} \\[6pt] & = \frac{ (2.3)^2 - (0.3)^2 }{(2.3)^2 -2 (2.3) (0.3)+ (0.3)^2} \\[6pt] & = \frac{ (2.3 + 0.3) (2.3 - 0.3) }{(2.3 - 0.3)^2 } \\[6pt] & = \frac{ (2.3 + 0.3) \cancel{(2.3 - 0.3)} }{(2.3 - 0.3) \cancel{(2.3 - 0.3)} } \\[6pt] &= \frac{2.3+0.3}{2.3-0.3} \\[6pt] &= \frac{2.6}{2} \\[6pt] &= 1.3 \end{aligned} \]

Answer \(\ \color{red}{1.3}\)

Solution

\[ \begin{aligned} 3x-4y &= 10 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] (3x-4y)^2 &= (10)^2 \\[6pt] 9x^2+16y^2-24xy &= 100 \\[6pt] 9x^2+16y^2-24 (-1) &= 100 \\[6pt] 9x^2+16y^2 + 24 &= 100 \\[6pt] 9x^2+16y^2 &= 100 - 24 \\[6pt] 9x^2+16y^2 &= 76 \\[6pt] \end{aligned} \]

Answer \( 9x^2+16y^2 = \color{red}{76}\)

Solution

\[ \begin{aligned} (5x-2y) &= 7 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] (5x-2y)^2 &= 7^2 \\[6pt] 25x^2 + 4y^2 - 20xy &= 49 \\[6pt] 25x^2 + 4y^2 - 20 \times 2 &= 49 \\[6pt] 25x^2 + 4y^2 - 40 &= 49 \\[6pt] 25x^2 + 4y^2 &= 49 + 40 \\[6pt] \color{green} 25x^2 + 4y^2 &= \color{green} 89 \\[8pt] (5x+2y)^2 &= 25x^2+4y^2+20xy \\[6pt] (5x+2y)^2 &= 89 + 20 \times 2 \\[6pt] (5x+2y)^2 &= 89 + 40 \\[6pt] (5x+2y)^2 &= 129 \\[6pt] \end{aligned} \]

Answer \( (5x+2y)^2 = \color{red}{129}\)

Solution

\[ \begin{aligned} \left(a+\frac{1}{a}\right) &= \frac{17}{4} \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(a+\frac{1}{a}\right)^2 &= \left(\frac{17}{4}\right)^2 \\[6pt] a^2+ \frac{1}{a^2} + 2 \times \cancel a \times \frac{1}{\cancel a} &= \frac{289}{16} \\[6pt] a^2+ \frac{1}{a^2} + 2 &= \frac{289}{16} \\[6pt] a^2+ \frac{1}{a^2} &= \frac{289}{16} - 2 \\[6pt] a^2+ \frac{1}{a^2} &= \frac{289 - 32}{16} \\[6pt] \color{green} a^2+ \frac{1}{a^2} &= \color{green} \frac{257}{16} \\[8pt] \left(a-\frac{1}{a}\right)^2 &= a^2+ \frac{1}{a^2} - 2 \times \cancel a \times \frac{1}{\cancel a} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= a^2+ \frac{1}{a^2} - 2 \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{257}{16} - 2 \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{257 - 32}{16} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{225}{16} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \left(\frac{15}{4}\right)^2 \\[6pt] a-\frac{1}{a} &= \frac{15}{4} \\[6pt] \end{aligned} \]

Answer \( \left(a-\dfrac{1}{a}\right) = \color{red}{\dfrac{15}{4}}\)