Hard Mathematical Induction I have a mathematical induction question and I know what I need to do just not how to do it.
The question is:

Prove the equality of:
$$(1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3$$

Base case:
$$(1 + 2)^2 = 1^3 + 2^3\\
(3)^2 = 1 + 8\\
9 = 9$$
and I understand I have to get the sides to equal each other though I'm not sure how to do that:
I use this:
$$(1 + . . . + n + (n + 1))^2 = 1^3 + . . . + n^3 + (n + 1)^3$$
but i can't seem to factor anything in anyway to figure it out . . .
I've tried putting the $S(n)$ in the $S(n + 1)$:
$$(1 + . . . + n + (n + 1))^2 = (1 + . . . + n)^2 + (n + 1)^3$$
but its just getting the $-(n + 1)^3$ on the first side I can't figure out...
Any help would be amazing!!!!

Let the statement hold for $n \in \mathbb{N}$. For $n+1$, by applying the binomial formula $(a+b)^2=a^2+2ab+b^2$ you have that $$\begin{align*}\left(\underbrace{1+2+\ldots+n}_{=a}+\underbrace{(n+1)}_{=b}\right)^2&=(1+2+\ldots+n)^2+2(1+2+\ldots+n)(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+2\frac{(n+1)n}{2}(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+\underbrace{n(n+1)^2+(n+1)^2}_{=(n+1)(n+1)^2=(n+1)^3}=\\&=1^3+2^3+\ldots+n^3+(n+1)^3 \end{align*}$$
