Basic induction proof methods so we're looking to prove $P(n)$ that 
$$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$
I know the basis step for $p(1)$ holds.
We're going to assume $P(k)$
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
And we're looking to prove $P(k+1)$
What I've discerned from the internet is that I should be looking to add the next term, $k+1$, to both sides so...
$$1^3+2^3+\cdots+k^3 + (k+1)^3=(k(k+1)/2)^2 + (k+1)^3$$
now I saw some nonsense since that we assumed $p(k)$ we can use it as a definition in our proof, specifically on the left hand side
so since 
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
then 
$$(k(k+1)/2)^2 + (k+1)^3 = (k(k+1)/2)^2 + (k+1)^3$$
and we have our proof
OK so far thats wrong
so far ive figured this.
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+1)+1)/2)^2$$
Then
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+2)/2)^2$$
using the definition
$$(k(k+1)/2)^2 + (k+1)^3 = ((k+1)((k+2)/2)^2$$
$$(k^2+k/2)^2 + (k^2+2k+1)(k+1) = (k^2+3k+2/2)^2$$
$$(k^4+k^2/4)+(k^2+2k^2+k+k^2+2k+1)= (k^4+9k^2+4/4)$$
Where should I go from here? It doesn't possibly look like these could equate, I'll keep going though

What you need to show is that $S(k-1)+k^3=S(k)$, i.e.
$$\frac{(k-1)^2k^2}4+k^3=\frac{k^2(k+1)^2}4.$$
Simplifying by $\frac{k^2}4$, you get
$$(k-1)^2+4k=(k+1)^2.$$
QED.
