Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
(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.
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