<s>[INST] Proof that sum of first $n$ cubes is always a perfect square I know that
$$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$
What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.
 [/INST] Let's prove that using the mathematical induction.
Check the base
$n = 1$
$1^3 = (\frac{1(1 + 1)}{2})^2$
That's true
Suppose that is true for $n - 1$ and prove that is true for $n$.
We know $1^3 + 2^3 + \dots + (n-1)^3 = (\frac{(n-1)n}{2})^2$
Then
$1^3 + 2^3 + \dots + (n-1)^3 + n^3 = (\frac{(n-1)n}{2})^2 + n^3 = \frac{n^2(n-1)^2}{4} + n^3 = \frac{n^4 - 2n^3 + n^2}{4} + n^3 = \frac{n^4 + 2n^3 + n^2}{4} = \frac{n^2(n+1)^2}{4} = (\frac{n(n+1)}{2})^2$
QED
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