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
 A: Assuming $P(k)$, you add $(k+1)^3$ on both sides of
$$
1^3 + 2^3 + \ldots + k^3 = (k(k+1)/2)^2
$$
to get
\begin{align}
1^3 + 2^3 + \ldots + k^3 + (k+1)^3
& = (k(k+1)/2)^2 + (k+1)^3 \\
& = \frac 14 k^2(k+1)^2 + (k+1)^3 \\
& = \frac 14\left(k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4\right) \\
& = \frac 14\left(k^4 + 6k^3 + 13k^2 + 12k + 4\right) \\
& = \frac 14(k+1)^2(k+2)^2 \\
& = \left((k+1)(k+2)/2\right)^2.
\end{align}
This statement is $P(k+1)$.
A: 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.
A: An example proof:
Let $P(k)$ denote the statement that $\sum\limits_{i = 1}^k i^3 = \left( \frac{k(k + 1)}{2} \right)^2$.
We wish to prove that for all $k \in \mathbb Z$ such that $k > 0$, $P(k)$.
We will prove this by induction on $k$.
In the base case, $k = 1$, and we have that $1^3 = 1 = \left(\frac{1(1 + 1)}{2}\right)^2$.
Now, for the inductive case, we suppose that for some $k$, $P(k)$ holds.
Then, as we have assumed $P(k)$, we have that $\sum\limits_{i = 1}^k i^3 = \left(\frac{k(k + 1)}{2} \right)^2$.
Adding $(k + 1)^3$ to both sides, we get that $\sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{k(k + 1)}{2} \right)^2 + (k + 1)^3$.
Now, the right-hand side is equal to, by factoring, $(k + 1)^2 \left(\frac{k^2}{4} + (k + 1)\right)$.
We can then rewrite this as $(k + 1)^2 \left(\frac{k^2 + (4(k + 1))}{4}\right)$, which then becomes $\frac{(k + 1)^2(k^2 + 4k + 4)}{4}$, or $\frac{(k + 1)^2(k + 2)^2}{4}$.
This is equal to $\left(\frac{(k + 1)(k + 2)}{2}\right)^2$, and so we have that
$$
\sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{(k + 1)(k + 2)}{2}\right)^2,
$$
which is precisely $P(k + 1)$.
Thus, we have shown that $P(1)$ holds and that for all $k \in \mathbb Z$ such that $k > 0$, $P(k) \implies P(k + 1)$, and so, by the principle of mathematical induction, $P(k)$ holds for all $k$.
