<s>[INST] Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$ This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$
$1^3 + 2^3 + ... + (k+1)^3$
$ = 1^3 + 2^3 + ... + k^3 + (k+1)^3$
$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.
Annnnd I'm stuck.  Not sure how to proceed from here on.
 [/INST] IMHO, this fact is a coincidence; a better approach is to prove the closed-form formula for both. As we know
$$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $$
the corresponding claim to prove is
$$ 1^3 + 2^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4} $$
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