Question
Which of the following are perfect cube$?\ 4608$
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Answer
On factorising $3087$ into prime factors, we get $4608 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$ Group the factors in triples of equal factors as: $4608 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 3 \times 3$ It is evident that the prime factors of $4608$ can be grouped into triples of equal factors and no factor is left over. Therefore, $4608$ is a perfect cube.
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