<s>[INST] 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.
 [/INST] Here's the idea. First note that your right hand side is actually $\frac{1}{4}N^2(N+1)^2$. Let's add $(N+1)^3$ to both sides of the inductive step to see if we can get $\frac{1}{4}(N+1)^2(N+2)^2$.
Doing this, we have
$$1^3 + 2^3 + \cdots + (N+1)^3 = \frac{1}{4}N^2(N+1)^2 + (N+1)^3 = (N+1)^2\left(\frac{1}{4}N^2 + N + 1\right) = \frac{1}{4}(N+1)^2(N^2+4N+4) = \frac{1}{4}(N+1)^2(N+2)^2.$$
From this, you clearly have that the truth of $P(N)$ implies the truth of $P(N+1)$. You should be able to handle write-up from here.
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