Show that every cube number is of the form $9 n$ or $9 n \pm 1$ .

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Solution

Every number $x$ is of from $3 p, 3 p \pm 1$
If $x = 3 p$
$\Rightarrow x^{3} = 27 p^{3} \Rightarrow x^{3} = 9 (3 p^{3})$
which is of form $9 n$
If $x = 3 p + 1$
$\Rightarrow x^{3} = {(3 p + 1)}^{3} \Rightarrow x^{3} = 27 p^{3} + 1 + 9 p (3 p + 1) \Rightarrow x^{3} = 27 p^{3} + 27 p + 9 p + 1$
which is of from $9 n + 1$
If $x = 3 p - 1$
$\Rightarrow x^{3} = {(3 p - 1)}^{3} \Rightarrow x^{3} = 27 p^{3} - 1 - 9 p (3 p - 1) \Rightarrow x^{3} = 27 p^{3} - 27 p + 9 p - 1$
which is of form $9 n - 1$
Hence proved.

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