Question
Is $128$ a perfect cube?
✓
Answer
By prime factorisation,
$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ [grouping the factors in triplets]
$=2^3 \times 2^3 \times 2$
In the above factorisation, $2$ remains after grouping the $2's$ in triplets.
Therefore, $128$ is not a perfect cube.
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