Prove that $(\sum_{i=1}^n i)^2$ = $\sum_{i=1}^n i^3$ by induction Prove that: $(\sum_{i=1}^n i)^2$ = $\sum_{i=1}^n i^3$
I can use the fact that $\sum_{i=1}^n i$ = n(n+1)/2 after the inductive hypothesis is invoked.
I'm not sure where to start, I would usually break down one side but there isn't usually two sums, so I'm not sure.

$$n=1 \to (\sum_{i=1}^1 i)^2=\sum_{i=1}^1 i^3 \to1=1\\
(***)n=k \to (\sum_{i=1}^k i)^2=\sum_{i=1}^k i^3\\
n=k+1 \to (\sum_{i=1}^{k+1}i)^2=\sum_{i=1}^{k+1} i^3\\
(\sum_{i=1}^{k+1}i)^2=(\sum_{i=1}^{k}i+(k+1))^2=(\sum_{i=1}^{k}i)^2+(k+1)^2+2(k+1)(\sum_{i=1}^{k}i)\to (***)\\=
\sum_{i=1}^k i^3+(k+1)^2+2(k+1)(\sum_{i=1}^{k}i)=\\
\sum_{i=1}^k i^3+(k+1)^2+2(k+1)(\dfrac{k(k+1)}{2})=\\
\sum_{i=1}^k i^3+(k+1)^2+(k+1)(k+1)k=\\
\sum_{i=1}^k i^3+(k+1)^2(1+k)=\\
\sum_{i=1}^k i^3+(k+1)^3=\\
\sum_{i=1}^{k+1} i^3$$
