Prove by induction: $\sum_{i=1}^n i^3 = \left[\sum_{i=1}^n i\right]^2$. Hint: Use $k(k+1)^2 = 2(k+1)\sum i$.
Basis: $n = 1$ $\sum_{i=1}^1 i^3 = \left[\sum_{i=1}^1 i\right]^2 \to 1^3 = 1^2 \to 1 = 1$.
Hypothesis: Assume true for all $n \le k$.
So far I have the following:
$$\sum_{i=1}^{k+1} i^3 = (k+1)^3 + \sum_{i=1}^k i^3$$
$$(k+1)^3 + \left[\sum_{i=1}^k i\right]^2$$
For $n=k+1$, $$\sum_{i=1}^{k+1}i^3 = \sum_{i=1}^{k}i^3+(k+1)^3=(\sum_{i=1}^{k}i)^2+(k+1)^3=(\sum_{i=1}^{k}i)^2+k(k+1)^2+(k+1)^2$$
Now using the Hint: $k(k+1)^2 = 2(k+1)\sum i$.
$$=(\sum_{i=1}^{k}i)^2+2(k+1)\sum_{i=1}^k i+(k+1)^2=(\sum_{i=1}^{k+1}i)^2$$