The sum of the cubes of the first \(n\) integers can be written as:
$$ 1^3 + 2^3 + 3^3 + ... + (n-1)^3 + n^3 = \frac{(n)^2(n+1)^2}{4} $$
Rearranging:
\begin{align} 1^3 + 2^3 + 3^3 + ... + (n-1)^3 &= \frac{(n)^2(n+1)^2}{4} - n^3 = \frac{(n)^2(n+1)^2 - 4n^3}{4} \\ &= \frac{(n)^2(n+1)^2 - n^2(4n)}{4} = \frac{(n)^2((n+1)^2 - 4n)}{4} \end{align}
If \([1^3 + 2^3 + 3^3 + ... + (n-1)^3 ≡ 0 \bmod n]\), then:
\begin{gather} \frac{(n)^2((n+1)^2 - 4n)}{4} ≡ 0 \bmod n \\ \frac{(n)^2((n+1)^2 - 4n)}{4} = nk \implies (n)((n+1)^2 - 4n) = 4k\end{gather}
This means \((n)((n+1)^2 - 4n)\) should be a multiple of 4, or \([(n)((n+1)^2 - 4n) ≡ 0 \bmod 4]\). There are four possible values for \(n\) modulo 4:
\begin{align} n &≡ 0 \bmod 4 \\ n &≡ 1 \\ n &≡ 2 \\ n &≡ 3 \end{align}
If \(n ≡ 1\):
\begin{gather} n+1 ≡ 2 ∧ 4n ≡ 0 \\ ((n)((n+1)^2 - 4n) ≡ (1)(2^2 - 0) ≡ 4 ≡ 0 \end{gather}
If \(n ≡ 2\):
\begin{gather} n+1 ≡ 3 ∧ 4n ≡ 0 \\ ((n)((n+1)^2 - 4n) ≡ (1)(3^2 - 0) ≡ 9 ≡ 1 \end{gather}
If \(n ≡ 3\):
\begin{gather} n+1 ≡ 0 ∧ 4n ≡ 0 \\ ((n)((n+1)^2 - 4n) ≡ (1)(0^2 - 0) ≡ 0 \end{gather}
This shows that only if \(n ≡ 2\), then \([(n)((n+1)^2 - 4n)]\) wouldn't be a multiple of 4. This means if \([1^3 + 2^3 + 3^3 + ... + (n-1)^3 ≡ 0 \bmod n]\), then \((n)((n+1)^2 - 4n)\) would be a multiple of 4, and so \(n \not ≡ 2\).