Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction. 
I have this: 
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1: 
And now what? I just couldn't solve it.

Perhaps the easiest way to go about this involves noticing that $$\frac 14N^4 + \frac 12N^3 + \frac 14N^2=\frac {N^2(N+1)^2}4$$
If you add $(N+1)^3$ to each side of $P(N)$ you get$$1^3+2^3+\dots N^3+(N+1)^3=\frac {N^2(N+1)^2}4+(N+1)^3$$
The left-hand side is what you need for $P(N+1)$ and the task is to show that the right-hand side can also be put into the correct form. You can extract an obvious common factor and put everything over a common denominator and see what happens.
If you don't happen to notice the convenient form for the right-hand side, you can still work with $\frac 14N^4 + \frac 12N^3 + \frac 14N^2$ to get the result.
