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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Updated on: 10-Oct-2022

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