Sponsor Area

Cubes And Cube Roots

Question
CBSEENMA8003436

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

Solution

(i) We have 243 = 3 x 3 x 3 x 3 x 3

The prime factor 3 is not a group of three.
∴ 243 is  not a perfect cube.
Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3  
Now, 729 becomes a [perfect cube
Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.
(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 256 in triples, we are left over with 2 x 2.
∴ 256 is  not a perfect cube.
Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2
or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
i.e. 512 is a perfect cube.
thus, the required smallest number is 2.
(iii) we  have 72  = 2 x 2 x 2 x 3 x 3

Grouping the prime factors of 72 in triples, we are left  over with 3 x 3
∴ 72 is  not a perfect cube.
Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3
or,     216 = 2 x 2 x 2 x 3 x 3 x 3
i.e. 216 is a perfect  cube
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.
(iv) We have 675 = 3 x 3 x 3 x 5 x 5
Grouping the prime factors of 675 to triples, we are left over with 5 x 5

∴  675 is not a perfect cube.
Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5
Now, 3375  is a  perfect cube
Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.
(v) We have 100 = 2 x 2 x 5 x 5
The prime factor are not in the groups of triples.

∴  100 is not a perfect cube.
Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5
or,   [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5
1000 = 2 x 2 x 2 x 5 x 5 x 5
Now, 1000 is a perfect cube
Thus, the required smallest number is 10