DAV Class 8 Maths Chapter 2 Worksheet 1
Cubes and Cube Roots Worksheet 1
1. Find the cubes of:
(i) 8
\[ \begin{align*} 8^3 &= 8 \times 8 \times 8 \\ &= 64 \times 8 \\ &= 512 \\ \end{align*} \]
Answer \( \color{red}8^3 = 512 \)
(ii) 13
\[ \begin{align*} (13)^3 &= 13 \times 13 \times 13 \\ &= 169 \times 13 \\ &= 2197 \\ \end{align*} \]
Answer \( \color{red}(13)^3 = 2197 \)
(iii) 17
\[ \begin{align*} 17^3 &= 17 \times 17 \times 17 \\ &= 289 \times 17 \\ &= 4913 \\ \end{align*} \]
Answer \( \color{red}(17)^3 = 4913 \)
(iv) 1.3
\[ \begin{align*} (1.3)^3 &= 1.3 \times 1.3 \times 1.3 \\ &= 1.69 \times 1.3 \\ &= 2.197 \\ \end{align*} \]
Answer \( \color{red}(1.3)^3 = 2.197 \)
(v) 0.06
\[ \begin{align*} (0.06)^3 &= 0.06 \times 0.06 \times 0.06 \\ &= 0.0036 \times 0.06 \\ &= 0.000216 \\ \end{align*} \]
Answer \( \color{red}(0.06)^3 = 0.000216 \)
(vi) 0.4
\[ \begin{align*} (0.4)^3 &= 0.4 \times 0.4 \times 0.4 \\ &= 0.16 \times 0.4 \\ &= 0.064 \\ \end{align*} \]
Answer \( \color{red}(0.4)^3 = 0.064 \)
(vii) \( \displaystyle \frac{2}{3} \)
\[ \begin{align*} \left(\frac{2}{3}\right)^3 &= \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\ \\ &= \frac{4}{9} \times \frac{2}{3} \\ \\ &= \frac{8}{27} \\ \end{align*} \]
Answer \( \displaystyle \color{red}\left(\frac{2}{3}\right)^3 = \frac{8}{27} \)
(viii) -7
\[ \begin{align*} (-7)^3 &= -7 \times -7 \times -7 \\ &= 49 \times -7 \\ &= -343 \\ \end{align*} \]
Answer \( \color{red}(-7)^3 = -343 \)
(ix) -9
\[ \begin{align*} (-9)^3 &= -9 \times -9 \times -9 \\ &= 81 \times -9 \\ &= -729 \\ \end{align*} \]
Answer \( \color{red}(-9)^3 = -729 \)
(x) -12
\[ \begin{align*} (-12)^3 &= -12 \times -12 \times -12 \\ &= 144 \times -12 \\ &= -1728 \\ \end{align*} \]
Answer \( \color{red}(-12)^3 = -1728 \)
2. Which of the following numbers are perfect cubes?
(i) 4096
\[ \begin{array}{c|c} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & \text{Prime factorisation of } 4096 \\ & \color{green}4096 = \boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \\ \end{aligned} \] Prime factors of 4096 can be grouped into triples and no factor is left over.
Answer \( \color{red} \text{4096 is a perfect cube} \)
(ii) 108
\[ \begin{array}{c|c} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & \text{Prime factorisation of } 108 \\ & 108 = 2 \times 2 \times \boxed{3 \times 3 \times 3} \end{aligned} \] While grouping the factors, we are left with \(2 \times 2 \)
Answer \( \color{red} \text{108 is a not perfect cube} \)
(iii) 392
\[ \begin{array}{c|c} 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & \text{Prime factorisation of } 392 \\ & 392 = \boxed{2 \times 2 \times 2} \times 7 \times 7 \end{aligned} \] While grouping the factors, we are left with \(7 \times 7 \)
Answer \( \color{red} \text{392 is not a perfect cube} \)
(iv) -27000
\[ \begin{array}{c|c} 2 & 27000 \\ \hline 2 & 13500 \\ \hline 2 & 6750 \\ \hline 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & \text{Prime factorisation of } 27000 \\ 27000 &= \boxed{2 \times 2 \times 2} \times \boxed{3 \times 3 \times 3} \times \boxed{5 \times 5 \times 5} \\ & = 2^3 \times 3^3 \times 5^3 \\ & = (2 \times 3 \times 5)^3 \\ 27000 & = (30)^3 \\ -27000 & = (-30)^3 \\ \end{aligned} \] Prime factors of -27000 can be grouped into triples and no factor is left over.
Answer \( \color{red} \text{-27000 is a perfect cube} \)
(v) \( \displaystyle \frac{-64}{1331} \)
\[ \begin{array}{cc} \begin{array}{c|c} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} & \quad \begin{array}{c|c} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array} \end{array} \] \[ \begin{aligned} \frac{-64}{1331} &= - \left(\frac{\boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2}} {\boxed{11 \times 11 \times 11}} \right) \\ \\ &= - \left(\frac{{2}^3 \times {2}^3} {{11}^3} \right) \\ \\ &= - \left(\frac{{4}^3} {{11}^3} \right) \\ \\ \frac{-64}{1331}&= \left(\frac{-4} {11} \right)^3 \\ \\ \end{aligned} \] Prime factors of \( \displaystyle \frac{-64}{1331} \) can be grouped into triples and no factor is left over.
Answer \( \displaystyle \color{red} \frac{-64}{1331} \text{ is a perfect cube} \)
3. Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.
Solution
\[ \begin{aligned} & \text{Prime factorisation of } 2560 \\ \end{aligned} \] \[ \begin{array}{c|c} 2 & 2560 \\ \hline 2 & 1280 \\ \hline 2 & 640 \\ \hline 2 & 320 \\ \hline 2 & 160 \\ \hline 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & 2560 = \boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \times 5 \\ \\ & \text{On grouping the factors we find out 5 is left out.} \\ & \text{So we need to multiply } 2560 \text{ by } 5 \times 5 = 25 \\ & \text{Required number } = 25 \\ \end{aligned} \]
Answer The smallest number required is \( \color{red} 25 \)
4. Find the smallest number by which 8788 must be divided so that the quotient is a perfect cube.
Solution
\[ \begin{aligned} & \text{Prime factorization of } 8788 \\ \end{aligned} \] \[ \begin{array}{c|c} 2 & 8788 \\ \hline 2 & 4394 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} & 8788 = 2 \times 2 \times \boxed{13 \times 13 \times 13} \\ \\ & \text{On grouping the factors we find out } 2 \times 2 \text{ is left out.} \\ & \text{So we need to divide } 8788 \text{ by } 2 \times 2 = 4 \\ & \text{Required number } = 4 \\ \end{aligned} \]
Answer The smallest number required is \( \color{red} 4 \)
5. Write true or false for the following statements:
(i) 650 is not a perfect cube.
Answer \( \color{red} True \)
\[ \begin{array}{c|c} 2 & 650 \\ \hline 5 & 325 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array} \] \[ \begin{aligned} 650 = 2 \times 5 \times 5 \times 13 \\ \\ \end{aligned} \]
(ii) Perfect cubes may end with two zeros.
Answer \( \color{red} False \)
\[ \begin{aligned} (10)^3 &= 10 \times 10 \times 10 \\ &= 1000 \\ \end{aligned} \]
(iii) Perfect cubes of odd numbers may not always be odd.
Answer \( \color{red} False \)
\[ \begin{aligned} (3)^3 &= 3 \times 3 \times 3 \\ &= 27 \\ (5)^3 &= 5 \times 5 \times 5 \\ &= 125 \\ \end{aligned} \]
(iv) Cube of negative numbers are negative.
Answer \( \color{red} True \)
\[ \begin{aligned} (-2)^3 &= -2 \times -2 \times -2 \\ &= 4 \times (-2) \\ &= -8 \\ \end{aligned} \]
(v) For a number to be a perfect cube, it must have prime factors in pairs.
Answer \( \color{red} True \)
\[ \begin{aligned} 8 &= 2 \times 2 \times 2 \\ 27 &= 3 \times 3 \times 3 \\ \end{aligned} \]