discrete mathematics - 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!!!!