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Find the cube root of each of the following numbers by prime factorization method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

= 2

= 2

iii) 10648 = 2 x 2 x 2 x 11 x 11 x 11

= 2

= 2

v) 15625 = 5 x 5 x 5 x 5 x 5 x 5

= 5

= 2

vii) 110592 = 2

= 2

State true or false.

(i) Cube of any odd number is even.

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

(iii) If 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.

i) As Odd multiplied by odd is always odd,So this statement is false

ii) A perfect cube will end with odd number of zeroes for example 10,the cube will be 1000.So this statement is true

iii) False

iv) As 2

v) The smallest two digit number is 10 and 10

vi) 99 is the largest 2 digit number; 99

vii) As 2

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

So for 1331

Left group 1

Right group 331

As you know 1

Now we have to find the cube root of left group 1

Now 1

So, we have 1 in ten’s place and 1 in unit place

11

Right group = 913

Left group = 4

7

We have to estimate the cube root of left group i.e 4

1

1<4<8

So, 10s digit in cube root of 4913 should be 1

So answer is 17

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**Notes**-
**Ncert Solutions**

Class 8 Maths Class 8 Science