Question
$999$ is a perfect cube.
✓
Answer
$\begin{array}{c|c}3 & 999 \\\hline3 & 333\\\hline3&111\\\hline37&37\\\hline&1\end{array}$
Resolving $999$ into prime factors, we get
$999 = 3 \times 3 \times 3 \times 37$
Grouping the factors in triplets of equal factors, we get
$999 = (3 \times 3 \times 3) \times 37$
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor $37$.
Therefore, $999$ is not a perfect cube.
Resolving $999$ into prime factors, we get
$999 = 3 \times 3 \times 3 \times 37$
Grouping the factors in triplets of equal factors, we get
$999 = (3 \times 3 \times 3) \times 37$
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor $37$.
Therefore, $999$ is not a perfect cube.
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