# NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots  Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 8 Maths Chapter 7 Cube And Cube Roots Exercise 7.1

1.Which of the following numbers are not perfect cubes?
(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

(i) We have the prime factors of 216 as:

$\begin{array}{cl}2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$216=2×2×2×3×3×3$

Thus, 216 is a perfect cube because in the factorisation, 2 and 3 have formed a group of three.

(ii) We have the prime factors of 128 as:

$\begin{array}{cl}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$128=2×2×2×2×2×2×2$

Here, 2 is left without making a group of three.So, 128 is not a perfect cube.

(iii) We have the prime factors of 1000 as:

$\begin{array}{cl}2& 1000\\ 2& 500\\ 2& 250\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$1000=2×2×2×5×5×5$

Thus, 1000 is a perfect cube because all the factors make perfect pairs.

(iv) We have the prime factors of 100 as:

$\begin{array}{cl}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$100=2×2×5×5$

Thus, 100 is not a perfect cube because 2 and 5 have not formed a required pair of three.

(v) We have the prime factors of 46656 as:

$\begin{array}{cl}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$46656=2×2×2×2×2×2×3×3×3×3×3×3$

Thus, 46656 is a perfect cube because all the factors are in perfect pairs of three.

2.Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

(i) We have the prime factors of 243 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$243=3×3×3×3×3={3}^{3}×3×3$

We can observe clearly that, number 3 is required to make $3×3$ a group of three. Thus, the required smallest number to be multiplied is 3.

(ii) We have the prime factors of 256 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$256=2×2×2×2×2×2×2×2={2}^{3}×{2}^{3}×2×2$

We can observe clearly that, number 2 is required to make $2×2$ a group of three.Thus, the required smallest number to be multiplied is 2.

(iii) We have the prime factors of 72 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$72=2×2×2×3×3={2}^{3}×3×3$

We can observe clearly that, number 3 is required to make $3×3$ a group of three.Thus, the required smallest number to be multiplied is 3.

(iv) We have the prime factors of 675 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$675=3×3×3×5×5={3}^{3}×5×5$

We can observe clearly that, number 5 is required to make $5×5$ a group of three.Thus, the required smallest number to be multiplied is 5.

(v) We have the prime factors of 100 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$243=2×2×5×5$

We can observe clearly that, number 2 and 5 are required to make $2×2$ and $5×5$ a group of three.Thus, the required smallest number to be multiplied is $5×2=10$.

3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81

(ii) 128

(iii) 135

(iv) 92

(v) 704

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

(i) We have the prime factors of 81 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$81=3×3×3×3={3}^{3}×3$

We can observe clearly that, number 3 is divided to 81 to make it a perfect cube.

So, 81 ÷ 3 = 27 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(ii) We have the prime factors of 256 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$128=2×2×2×2×2×2×2={2}^{3}×{2}^{3}×2$

We can observe clearly that, number 2 is divided to 128 to make it a perfect cube.

So, 128 ÷ 2 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 2.

(iii) We have the prime factors of 135 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$

$81=3×3×3×5={3}^{3}×5$

We can observe clearly that, number 5 is divided to 135 to make it a perfect cube.

So, 135 ÷ 5= 27 which is a perfect cube. Thus, the required smallest number to be divided is 5.

(iv) We have the prime factors of 192 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$

$192=2×2×2×2×2×2×3={2}^{3}×{2}^{3}×3$

We can observe clearly that, number 3 is divided to 192 to make it a perfect cube.

So, 192 ÷ 3 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(v) We have the prime factors of 704 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 704\\ 2& 352\\ 2& 176\\ 2& 88\\ 2& 44\\ 2& 22\\ 11& 11\\ & 1\end{array}$

$128=2×2×2×2×2×2×11={2}^{3}×{2}^{3}×11$

We can observe clearly that, number 11 is divided to 704 to make it a perfect cube.

So, 704 ÷ 11 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 11.

4.Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

Given, the sides of the cuboid as 5 cm, 2 cm and 5 cm.

So we have the volume of the cuboid =$5cm×2cm×5cm=50c{m}^{3}$

We have the prime factors of 50 as:

$50=2×5×5$

To make it a perfect cube, we must have $2×2×2×5×5×5$

$?20×\left(2×5×5\right)$

$?20×volume\phantom{\rule{0.278em}{0ex}}of\phantom{\rule{0.278em}{0ex}}the\phantom{\rule{0.278em}{0ex}}given\phantom{\rule{0.278em}{0ex}}cuboid$

Thus,we have the required number of cuboids = 20.

## NCERT solutions for class 8 Maths Chapter 7 Cube And Cube Roots Exercise 7.2

1.Find the cube root of each of the following numbers by prime factorisation method.
(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

(i) We have the prime factors of 64 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$256=2×2×2×2×2×2={2}^{3}×{2}^{3}$

?$\sqrt{64}=2×2=4$

Hence, the cube root of 64 is 4.

(ii) We have the prime factors of 512 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$256=2×2×2×2×2×2×2×2×2={2}^{3}×{2}^{3}×{2}^{3}$

?$\sqrt{512}=2×2×2=8$

Hence, the cube root of 512 is 8.

(iii) We have the prime factors of 10648 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 10648\\ 2& 5324\\ 2& 2662\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}$

$10648=2×2×2×11×11×11={2}^{3}×{11}^{3}×$

?$\sqrt{10648}=2×11=22$

Hence, the cube root of 10648 is 22.

(iv) We have the prime factors of 27000 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 27000\\ 2& 13500\\ 2& 6750\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$27000=2×2×2×3×3×3×5×5×5={2}^{3}×{3}^{3}×{5}^{3}$

?$\sqrt{27000}=2×3×5=30$

Hence, the cube root of 27000 is 30.

(v) We have the prime factors of 15625 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}5& 15625\\ 5& 3125\\ 5& 625\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$15625=5×5×5×5×5×5={5}^{3}×{5}^{3}$

?$\sqrt{15625}=5×5=25$

Hence, the cube root of 15625 is 25.

(vi) We have the prime factors of 13824 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$13824=2×2×2×2×2×2×2×2×2×3×3×3$

$?{2}^{3}×{2}^{3}×{2}^{3}×{3}^{3}$

?$\sqrt{13824}=2×2×2×3=24$

Hence, the cube root of 13824 is 24.

(vii) We have the prime factors of 110592 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 110592\\ 2& 55296\\ 2& 27648\\ 2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$110592=2×2×2×2×2×2×2×2×2×2×2×2×3×3×3$

$?{2}^{3}×{2}^{3}×{2}^{3}×{2}^{3}×{3}^{3}$

?$\sqrt{110592}=2×2×2×2×3=48$

Hence, the cube root of 110592 is 48.

(viii) We have the prime factors of 46656 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$46656=2×2×2×2×2×2×3×3×3×3×3×3$

$?{2}^{3}×{2}^{3}×{3}^{3}×{3}^{3}$

?$\sqrt{46656}=2×2×3×3=36$

Hence, the cube root of 46656 is 36.

(ix) We have the prime factors of 175616 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}2& 175616\\ 2& 87808\\ 2& 43904\\ 2& 21952\\ 2& 10976\\ 2& 5488\\ 2& 2744\\ 2& 1372\\ 2& 686\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$175616=2×2×2×2×2×2×2×2×2×7×7×7$

$?{2}^{3}×{2}^{3}×{2}^{3}×{7}^{3}$

?$\sqrt{175616}=2×2×2×7=56$

Hence, the cube root of 175616 is 56.

(x) We have the prime factors of 91125 as:

$\phantom{\rule{2em}{0ex}}\begin{array}{cl}3& 91125\\ 3& 30375\\ 3& 10125\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}$

$91125=3×3×3×3×3×3×5×5×5={3}^{3}×{3}^{3}×{5}^{3}$

?$\sqrt{91125}=3×3×5=45$

Hence, the cube root of 91125 is 45.

2.State True or False.
(i) Cube of an odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If the square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

(i) False because Cube of any odd number is always odd, e.g., $\left(3{\right)}^{3}=27$

(ii) True.

(iii) True. For e.g., $\left(5{\right)}^{2}=25$ and $\left(5{\right)}^{3}=625$

(iv) False because there exists perfect cube which ends with 8 for e.g., $\left(12{\right)}^{3}=1728$ending with 8

(v) False. let's take the lowest two digit number i.e., $\left(10{\right)}^{3}=1000$which is a 4-digit number.

(vi) False.Let's take the highest two digit number i.e., $\left(99{\right)}^{3}=970299$ which is 6-digit number.

(vii) True. For e.g., $\left(2{\right)}^{3}=8$ which is 1-digit number.

3.You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

We have the perfect cube = 1331
Now let us form groups of three from the rightmost digits of 133

So we have,

Group 2= 1 and, group 1= 331

One’s digit in first group = 1, that says that one’s digit in the required cube root may be 1.Also, in the second group we have only 1.

Estimated cube root of 1331 = 11

Thus, $\sqrt{1331}=11$

(i) We have the perfect cube = 4913

Now let us form groups of three from the rightmost digits of 4913

So we have,

group 2 = 4 And,group 1= 913

One’s place digit in 913 is 3.So that says that one’s place digit in the cube root of the given number may be 7.

Now in group 2 digit is 4 and we know that 4 lies as:${1}^{3}<4<{2}^{3}$.We know that, ten’s place must be the smallest number i.e,1.

Thus, the estimated cube root of 4913 = 17.

(ii) We have the perfect cube = 12167

Now let us form groups of three from the rightmost digits of 12167

So we have,

group 2= 12 and, group 1= 167

The ones place digit in 167 is 7.So we can say that one’s digit in the cube root of the given number may be 3.

Now in group 2, we have 12 and 12 lies between as:${2}^{3}<12<{3}^{3}$. Also, we know that the ten’s place of the required cube root of the given number = 2

. Thus, the estimated cube root of 12167 = 23.

(iii) We have the perfect cube = 32768

Now let us form groups of three from the rightmost digits of 32768, we have

So we have,

group 2= 32 and, group 1= 768

One’s place digit in 768 is 8.So we can say that the one’s place digit in the cube root of the given number may be 2.

Now in group 2, we have 32 and 32 lies between the numbers as:${3}^{3}<32<{4}^{3}$.Also, we know that the ten’s place of the cube root of the given number = 3.

Thus, the estimated cube root of 32768 = 32.

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