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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.