I'm having trouble with induction. Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ I started a new course and I'm expected to know this stuff, and I'm having trouble learning some on my own.
I'm stuck with this problem:
Prove  $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ using induction (the $1+2^3...$ is written in sum notation, although I don't know how to enter that here, sorry).
I started substituting $n \rightarrow n+1$ but I don't know what to do next. Any help would be extremely appreciated.

This question often causes a lot of confusion simply because of the cumbersome factoring. 
Let $P(n)$ be the proposition that $$\sum_{i=1}^n i^3 = \frac{(n^2)(n+1)^2}{4}.$$
Base Case, $P(1)$: $\sum_{i=1}^1 i^3 = 1$ and $\frac{(1^2)(1+1)^2}{4} = \frac{4}{4} = 1$ so $P(1)$ is true. 
Inductive Hypothesis: Assume that $P(k)$ is true: $1^3 + \cdots + k^3 = \frac{(k^2)(k+1)^2}{4}$
Now it remains to show that $P(k) \implies P(k+1)$. Start with the left side of the equation: 
$$\sum_{i=1}^{k+1} i^3 = \left( \sum_{i=1}^k i^3 \right) + (k+1)^3 = \frac{(k^2)(k+1)^2}{4} + (k+1)^3$$
(by the IH above). Now we factor out a $(k+1)^2$: $$= (k+1)^2 \cdot (\frac{k^2}{4} + k + 1) = \frac{(k+1)^2(k^2+4k+4)}{4} = \frac{(k+1)^2((k+1)+1)^2}{4}$$
Which is equal to the desired result. 
