Cube and cube root Solved for class 8

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

2. State true or false.

(i) Cube of any odd number is even.

FALSE: Odd multiplied by odd is always odd

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

TRUE: A perfect cube will end with odd number of zeroes

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

TRUE: 5 multiplied by 5 any number of times always gives 5 at unit’s place

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

False: 2³ = 8

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

FALSE: The smallest two digit number is 10 and 10³ = 1000 is a three digit number

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

FALSE: 99 is the largest 2 digit number; 99³ = 989901 is a 6 digit number

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

TRUE: 2³ = 8 is a single 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 root of 4913.

Answer: Let us divide 1331 in two groups of 31 and 13 for extreme right half and extreme left half of the 
number.

As you know 1³ = 1 so there would be 1 at unit’s place in cube root of 1331.

Now 2³ = 8 and 3³ = 27

It is clear that 8<13<27, so the 10s digit of cube root of 1331 may be 2

So, cube root of 1331 may be 21 but 21³ = 9261 is not equal to 1331

So, let us test the 10s digit as 1

11³ = 1331 satisfies the condition

4913:

Right group = 13

Left group = 49

7³ gives 3 at unit’s place so unit digit number in cube root of 4913 should be 7

3³ = 27 and 4³ = 64

27<49<64

So, 10s digit in cube root of 4913 should be 3

Test: 37³ = 50653 is not equal to 4913

Let Us test 27³ = 19683 ¹ 4913

Let us test 17³ = 4913 gives the answer