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.