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# Which of the following numbers are not perfect cubes?

**(i)** 64

**(ii) **216

**(iii)** 243

**(iv) **1728.

To do:

We have to find the whether the given numbers are perfect cubes.

Solution:

(i) Prime factorisation of 64 is,

$64=2\times2\times2\times2\times2\times2$

$=2^3\times2^3$

$=(2\times2)^3$

$=4^3$

Grouping the factors in triplets of equal factors, we see that no factor is left.

Therefore,

64 is a perfect cube.

(ii) Prime factorisation of 216 is,

$216=2\times2\times2\times3\times3\times3$

$=2^3\times3^3$

$=(2\times3)^3$

$=6^3$

Grouping the factors in triplets of equal factors, we see that no factor is left.

Therefore,

216 is a perfect cube.

(iii) Prime factorisation of 243 is,

$243=3\times3\times3\times3\times3$

$=3^3\times3^2$

Grouping the factors in triplets of equal factors, we see that two factors $3 \times 3$ are left.

Therefore,

243 is not a perfect cube.

(iv) Prime factorisation of 1728 is,

$1728=2\times2\times2\times2\times2\times2\times3\times3\times3$

$=2^3\times2^3\times3^3$

Grouping the factors in triplets of equal factors, we see that no factor is left.

Therefore,

1728 is a perfect cube.

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