Which of the following numbers are not perfect cubes?

Question:

Which of the following numbers are not perfect cubes?

(i) 64

(ii) 216

(iii) 243

(iv) 1728

Solution:

(i)
On factorising 64 into prime factors, we get:

$64=2 \times 2 \times 2 \times 2 \times 2 \times 2$

On grouping the factors in triples of equal factors, we get:

$64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$

It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.

(ii)
On factorising 216 into prime factors, we get:

$216=2 \times 2 \times 2 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$216=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$

It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.

(iii)
On factorising 243 into prime factors, we get:

$243=3 \times 3 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$243=\{3 \times 3 \times 3\} \times 3 \times 3$

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube

(iv)
On factorising 1728 into prime factors, we get

$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$1728=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$

It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.

Thus, (iii) 243 is the required number, which is not a perfect cube.