algebra precalculus - Prove that $(1+2+3+cdots+n)^2=1^3+2^3+3^3+cdots+n^3$ $forall n in mathbb{N}$.

Prove that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ for every $n \in \mathbb{N}$.

I'm trying to use induction on this one, but I'm not sure how to. The base case is clearly true. But when I add $n+1$ to the right and left hand sides, I don't know what I'm supposed to get. For example, when I extend the right hand side's sequence by $n+1$, I get $n^3+(n+1)^3$ at the end of the sequence, which is $2n^3 +3n^2+3n+1$.

This doesn't seem that meaningful, so I don't know how to proceed.