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Find the smallest number by which each of the numbers must be multiplied to obtain a perfect cube

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Answer:

A number is a perfect cube only when each factor in the prime factorization of the given number exists in triplets. Using this concept, the smallest number can be identified.

Explanation:

for example:-

we take ,

i)243

ii)256

ii.72

iv.675

v.100

(i) 243

243 = 3 × 3 × 3 × 3 × 3

= 33 × 32

Here, one group of 3's is not existing as a triplet. To make it a triplet, we need to multiply by 3.

Thus, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube

Hence, the smallest natural number by which 243 should be multiplied to make a perfect cube is 3.

(ii)

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 23 × 23 × 2 × 2

Here, one of the groups of 2’s is not a triplet. To make it a triplet, we need to multiply by 2.

Thus, 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube

Hence, the smallest natural number by which 256 should be multiplied to make a perfect cube is 2.

(iii) 72

72 = 2 × 2 × 2 × 3 × 3

= 23 × 32

Here, the group of 3’s is not a triplet. To make it a triplet, we need to multiply by 3.

Thus, 72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube

Hence, the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.

(iv) 675

675 = 5 × 5 × 3 × 3 × 3

= 52 × 33

Here, the group of 5’s is not a triplet. To make it a triplet, we need to multiply by 5.

Thus, 675 × 5 = 5 × 5 × 5 × 3 × 3 × 3 = 3375 is a perfect cube

Hence, the smallest natural number by which 675 should be multiplied to make a perfect cube is 5.

(v) 100

100 = 2 × 2 × 5 × 5

= 22 × 52

Here both the prime factors are not triplets. To make them triplets, we need to multiply by one 2 and one 5.

Thus, 100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make a perfect cube is 2 × 5 =10

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