Question

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube: 

1. 81 
2. 128 
3.  135 
4.  192 
5.  704

Answer 1

(i) \(81\)

Prime factors of \(81\) =  \(3\times3\times3\times3\times\)
Here one factor \(3\) is not grouped in triplets.
Therefore \(81\) must be divided by \(3\) to make it a perfect cube.

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(ii) \(128\)

Prime factors of \(128\) = \(2\times2\times2\times2\times2\times2\times2\times\)
Here one factor \(2 \) does not appear in a \(3’\)s group.
Therefore, \(128\) must be divided by \(2\) to make it a perfect cube.

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(iii) \(135\)

Prime factors of \(135\) = \(3\times3\times3\times5\)
Here one factor \(5\) does not appear in a triplet.
Therefore, \(135\) must be divided by \(5\) to make it a perfect cube.

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(iv) \(192\)

Prime factors of \(192\) = \(2\times2\times2\times2\times2\times2\times3\)
Here one factor \(3\) does not appear in a triplet.
Therefore, \(192\) must be divided by \(3\) to make it a perfect cube.

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(v) \(704\)

Prime factors of \(704\) = \(2\times2\times2\times2\times2\times2\times11\)
Here one factor \(11\) does not appear in a triplet.
Therefore, \(704\) must be divided by \(11\) to make it a perfect cube.