# Selina Concise Class 8 Math Chapter 4 Cubes And Cube-roots Exercise 4A Solutions

## EXERCISE 4A

(1) Find the cube of:

(i) 73 = 7 × 7 × 7 = 343

(ii) (11)3 = 11 × 11 × 11 = 1331

(iii) (16)3 = 16 × 16 × 16 = 4096

(iv) (23)3 = 23 × 23 × 23 = 12167

(v) (31)3 = 31 × 31 × 31 = 29791

(vi) (42)3 = 42 × 42 × 42 = 74088

(vii) (54)3 = 54 × 54 × 54 = 157464

(2) Find which of the following are perfect cubes?

(i) 243 = 7 × 7 × 7 =  (7)3

∴ 243 is a perfect cube.   (3) Find the cubes of:

(i) 2.1 = (2.1)3 = 2.1 × 2.1 × 2.1 = 9.261

(ii) 0.4 = (0.4)3 = 0.4 × 0.4 × 0.4 = 0.064

(iii) 1.6 = (1.6)3 = 1.6 × 1.6 × 1.6 = 4.096

(iv) 2.5 = (2.5)3 = 2.5 × 2.5 × 2.5 = 15.625

(v) 0.12 = (0.12)3 = 0.12 × 0.12 × 0.12 = 0.001728

(vi) 0.02 = (0.02)3 = 0.02 × 0.02 × 0.02 = 0.000008

(vi) 0.8 = (0.8)3 = 0.8 × 0.8 × 0.8 = 0.512

(4) Find the cubes of: (5) Find the cubes of:

(i) (- 3)3 = – 3 × – 3 × -3 = – 27

(ii) (- 7)3 = – 7 × – 7 × – 7 = – 343

(iii) (- 12)3 = – 12 × – 12 × – 12 = – 1728

(iv) (- 18)3 = – 18 × – 18 × – 18 = – 5832

(v) (- 25)3 = – 25 × – 25 × – 25 = – 15625

(vi) (- 30)3 = – 30 × – 30 × – 30 = – 27000

(vii) (- 50)3 = – 50 × – 50 × – 50 = – 125000

(6) Which of the following are cubes of:  (7) The prime factor of 1323 are = 3 × 3 × 3 × 7 × 7

Clearly, 1323 must be multiplied by 7.  Updated: April 2, 2019 — 3:34 pm

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