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.
When you extend the right side you don't get $n+(n+1)^3$, you get $\sum_{i=1}^ni^3+(n+1)^3$ In the original question it was $n$, but in the edit it was $n^3$. Both are incorrect.