SEBA Class 10 Maths Chapter R-2 Solutions: Cube and Cube Root | Revision Chapter

Get free SEBA Class 10 Maths Revision Exercise R-2 Solutions: Cube and Cube Root based on the new SCERT textbook 2026. In this article, we have solved all questions of this revision chapter, R-2, in a simple, step-by-step manner. 

See More:

• SEBA Class 10 Maths Chapter R-1 Solutions: Square and Square Root
• SEBA Class 10 Maths Chapter R-2 Solutions: Cube and Cube Root
• SEBA Class 10 Maths Chapter R-3 Solutions: Indices and Power
• SEBA Class 10 Maths Chapter R-4 Solutions: Factorisation
• SEBA Class 10 Maths Chapter R-5 Solutions: Congruence of Triangles

1. Which of the following is not a perfect cube?

(i) 3757

(ii) 3375

(iii) 3332

(iv) 4096

$\mathbf{Sol^n.}$

A number is a perfect cube if, in its prime factorisation, each prime factor occurs in groups of three (as a triplet).

• For (i) 3757:

  $3757 = 11 \times 11 \times 31 $
  $= 11^2 \times 31$

  Here, neither $11$ nor $31$ forms a triplet.

  $\therefore 3757$ is not a perfect cube.

• For (ii) 3375:

  $3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 $
  $= 3^3 \times 5^3 $
  $= (3 \times 5)^3 $
  $= 15^3$

  $\therefore 3375$ is a perfect cube.

• For (iii) 3332:

  $3332 = 2 \times 2 \times 7 \times 7 \times 17 $
  $= 2^2 \times 7^2 \times 17$

  Here, the factors do not occur in triplets.

  $\therefore 3332$ is not a perfect cube.

• For (iv) 4096:

  $4096 = 2^{12} $
  $= (2^4)^3 $
  $= 16^3$

  $\therefore 4096$ is a perfect cube.

Therefore, (i) 3757 and (iii) 3332 are not perfect cubes.

2. Find the cubes of the following numbers.

(i) 19

$\mathbf{Sol^n.}$

$19^3 = 19 \times 19 \times 19$

$= 361 \times 19$

$= 6859$

(ii) 21

$\mathbf{Sol^n.}$

$21^3 = 21 \times 21 \times 21$

$= 441 \times 21$

$= 9261$

(iii) 23

$\mathbf{Sol^n.}$

$23^3 = 23 \times 23 \times 23$

$= 529 \times 23$

$= 12167$

(iv) 27

$\mathbf{Sol^n.}$

$27^3 = 27 \times 27 \times 27$

$= 729 \times 27$

$= 19683$

3. Write the digit in the unit place of the cubes of the following numbers.

(i) 14

$\mathbf{Sol^n.}$

Unit digit is $4$.

$4^3 = 64$ (unit digit is $4$)

Therefore, the digit in the unit place of the cube of $14$ is $4$.

(ii) 18

Unit digit is $8$.

$8^3 = 512$ (unit digit is $2$)

Therefore, the digit in the unit place of the cube of $18$ is $2$.

(iii) 13

Unit digit is $3$.

$3^3 = 27$ (unit digit is $7$)

Therefore, the digit in the unit place of the cube of $13$ is $7$.

(iv) 27

Unit digit is $7$.

$7^3 = 343$ (unit digit is $3$)

Therefore, the digit in the unit place of the cube of $27$ is $3$.

4. Find the smallest integers with which the following numbers are to be multiplied so that they become perfect cubes.

(i) 5324

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$5324 = 2 \times 2 \times 11 \times 11 \times 11 $
$= 2^2 \times 11^3$

Here, the prime factor $2$ does not appear in a group of three.

To make it a perfect cube, it must be multiplied by one more $2$.

(ii) 3087

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$3087 = 3 \times 3 \times 7 \times 7 \times 7 $
$= 3^2 \times 7^3$

Here, the prime factor $3$ does not appear in a triplet.

Hence, the smallest integer to be multiplied is $3$.

(iii) 3125

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$3125 = 5 \times 5 \times 5 \times 5 \times 5 $

$= 5^3 \times 5^2$

So, the smallest integer to be multiplied is $5$.

(iv) 648

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 $

$= 2^3 \times 3^3 \times 3$

So, the smallest integer to be multiplied is $3 \times 3 =$ $9$.

5. Find the smallest numbers with which the following numbers are to be divided so that they become perfect cube.

(i) 10,368

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$10368 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= 2^7 \times 3^3$

$= (2^3) \times (2^3) \times 2 \times (3^3)$

So, the smallest number to be divided is $2$.

(ii) 2187

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$2187 $

$= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^7$

$= (3^3) \times (3^3) \times 3$

Hence, the smallest number to be divided is $3$.

(iii) 5000

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$5000 $

$= 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 $

$= 2^3 \times 5^4$

$= (2^3) \times (5^3) \times 5$

Therefore, the smallest number to be divided is $5$.

(iv) 8192

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$8192 = 2^{13}$

$= (2^3) \times (2^3) \times (2^3) \times (2^3) \times 2$

Therefore, the smallest number to be divided is $2$.

6. Find the cube roots of the following numbers.

(i) 1331

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$1331 = 11 \times 11 \times 11 = 11^3$

$\therefore \sqrt[3]{1331} = 11$

(ii) 1728

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$1728 $

$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$= 2^3 \times 2^3 \times 3^3$

$= (2 \times 2 \times 3)^3 = 12^3$

$\therefore \sqrt[3]{1728} = 12$

(iii) 2197

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$2197 = 13 \times 13 \times 13 $

$= 13^3$

$\therefore \sqrt[3]{2197} = 13$

(iv) 2744

$\mathbf{Sol^n.}$

By prime factorisation, we get:

$2744 $

$= 2 \times 2 \times 2 \times 7 \times 7 \times 7$

$= 2^3 \times 7^3$

$= (2 \times 7)^3 = 14^3$

$\therefore \sqrt[3]{2744} = 14$

7. Multiple Choice Questions

(a) The digit in the unit place in the cube of 23 is

(i) 6

(ii) 7

(iii) 8

(iv) 9

$\mathbf{Sol^n.}$

Unit digit of $23$ is $3$.

Cube of $3 = 3^3 = 27$, which ends in $7$.

$\therefore$ So the correct option is (ii) 7

(b) Which of the following is a perfect cube?

(i) 652

(ii) 933

(iii) 343

(iv) 1002

$\mathbf{Sol^n.}$

We know that:

$7 \times 7 \times 7 = 343 = 7^3$

$\therefore 343$ is a perfect cube.

$\therefore$ So the correct option is (iii) 343

(c) The value of $\sqrt[3]{1000}$ is

(i) 30

(ii) 100

(iii) 10

(iv) 1000

$\mathbf{Sol^n.}$

Since $10^3 = 1000$, we have $\sqrt[3]{1000} = 10$.

$\therefore$ So the correct option is (iii) 10

(d) If $m$ is the cube root of $n$ then the value of $n$ is

(i) $\sqrt{m}$

(ii) $\sqrt[3]{m}$

(iii) $m^3$

(iv) $m^2$

$\mathbf{Sol^n.}$

Given: $\sqrt[3]{n} = m$

Cubing both sides:

$(\sqrt[3]{n})^3 = m^3 $

$\implies n = m^3$

$\therefore$ Correct Option: (iii) $m^3$

(e) The value of $\sqrt[3]{8} + \sqrt[3]{27} + \sqrt[3]{64}$ is

(i) 6

(ii) 7

(iii) 8

(iv) 9

$\mathbf{Sol^n.}$

$\sqrt[3]{8} = \sqrt[3]{2^3} = 2$

$\sqrt[3]{27} = \sqrt[3]{3^3} = 3$

$\sqrt[3]{64} = \sqrt[3]{4^3} = 4$

Therefore:

$\sqrt[3]{8} + \sqrt[3]{27} + \sqrt[3]{64} $

$= 2 + 3 + 4 = 9$

$\therefore$ So the correct option is (iv) 9