Which of the following are perfect cubes?
(i) 64
(ii)216
(iii) 243
(iv) 1000
(v) 1728
(vi) 3087
(vii) 4608
(viii) 106480
(ix) 166375
(x) 456533.

To do:

We have to find the whether the given number is a perfect cube.

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 1000 is,

$1000=2\times2\times2\times5\times5\times5$

$=2^3\times5^3$

$=(2\times5)^3$

$=15^3$

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

Therefore,

1000 is a perfect cube.

(v)Prime factorisation of 1728 is,

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

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

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

$=12^3$

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

Therefore,

1728 is a perfect cube.

(vi) Prime factorisation of 3087 is,

$3087=3\times3\times7\times7\times7$

$=3^2\times7^3$

Grouping the factors in triplets of equal factors, we see that two factor are left.

Therefore,

3087 is not a perfect cube.

(vii) Prime factorisation of 4608 is,

$4608=2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3$

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

Grouping the factors in triplets of equal factors, we see that two factor are left.

Therefore,

4608 is not a perfect cube.

(viii) Prime factorisation of 106480 is,

$106480=2\times2\times2\times2\times5\times11\times11\times11$

$=2^3\times2\times5\times11^3$

Grouping the factors in triplets of equal factors, we see that two factors(2 and 5) are left.

Therefore,

106480 is not a perfect cube.

(ix) Prime factorisation of 166375 is,

$166375=5\times5\times5\times11\times11\times11$

$=5^3\times11^3$

$=(5\times11)^3$

$=55^3$

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

Therefore,

166375 is a perfect cube.

(x) Prime factorisation of 456533 is,

$456533=7\times7\times7\times11\times11\times11$

$=7^3\times11^3$

$=(7\times11)^3$

$=77^3$

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

Therefore,

456533 is a perfect cube.

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Updated on: 10-Oct-2022

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