Prove $\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$ (induction) Problem
Prove $$\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$$
 Attempt to solve 
I would try to prove this with induction. We have sum and the sum as function of $p(n)$. Now i try to prove that the sum equals $p(n)$ with induction. $p(n)=\frac{1}{4}n^2(n+1)^2$
$$ 1^3+2^3+\cdots + \ n^3 = \frac{1}{4}n^2(n+1)^2 $$
Base case
$$ p(1)=\frac{1}{4}\cdot 1^2 (1+1)^2 = \frac{2^2}{4} = \frac{4}{4} = 1 $$
Induction step
Assume that $n \ge 1$ and $p(n)$ is valid. 
$$ 1^3+2^3+ \dots + n^3 = \frac{1}{4} n^2(n+1)^2$$
substitute $n=n+1$
$$ 1^3+2^3 + \dots + n^3 + (n+1)^3 = \frac{1}{4} (n+1)^2 ((n+1)+1)^2 $$
$$ 1^3+2^3 + \dots + n^3 + (n+1)^3 = \frac{1}{4} (n+1)^2 (n+2)^2 $$
we know that 
$$ 1^3+2^3+ \dots + n^3 = \frac{1}{4} n^2(n+1)^2$$
so we have
$$ \frac{1}{4}n^2(n+1)^2 + (n+1)^3 = \frac{1}{4}n^2(n+1)^2(n+2)^2 || \cdot 4 $$
$$ n^2(n+1)^2+4(n+1)^3=n^2(n+1)^2(n+2)^2 $$
$$ n^2(n+1)^2+4(n+1)^2(n+1)=n^2(n+1)^2(n+2)^2 || : (n+1)^2$$
$$ n^2+4(n+1)=n^2(n+2)^2 $$
$$ n^2+4n+4  = n^2(n^2+4n+4)$$
$$ n^2+4n +4 \neq n^2(n^2+4n+4) $$
Problem is there is extra $n^2$ on right side compared to left side which makes equation not true. Is there computational error or is the problem more fundamental perhaps i don't understand induction proofs well enough?

As @packsciences said, your mistake is after "so we have". The correct way to do it is:
$$\frac{1}{4}n^2(n+1)^2 + (n+1)^3 = (n+1)^2 (\frac{1}{4}n^2 + n+1) = \frac{1}{4}(n+1)^2 (n^2 + 4n + 4) = \frac{1}{4} (n+1)^2 (n+2)^2$$
