Your post is extremely hard to read. What I think you are trying to prove is:

$\displaystyle \sum_{i = 1}^n i^3 = \dfrac{n^2(n+1)^2}{4}$

It is clear this holds for $n = 1$.

Assuming that:

$\displaystyle \sum_{i = 1}^k i^3 = \dfrac{k^2(k+1)^2}{4}$ for a given $k$, we have:

$\displaystyle \sum_{i = 1}^{k+1} i^3 = \sum_{i = 1}^k i^3 + (k+1)^3 = \dfrac{k^2(k+1)^2}{4} + (k+1)^3$

$= \dfrac{(k+1)^2}{4}[k^2 + 4(k+1)]$

and you may continue from here. You are not supposed to just "replace" $k$ with $k+1$ and see if it works.