I'm using sigma notation and this is what it breaks down to:
p(k) is k^3= (k^2(k+1)^2)/4
Are you asking how to show that
$\displaystyle \sum_{i = 1}^n i^3 = \frac{1}{4} \cdot n^2(n+ 1)^2$?
The base case is n = 1:
$\displaystyle \sum_{i = 1}^1 1^3 = 1$
and
$\displaystyle \frac{1}{4} \cdot 1^2(1 + 1)^2 = \frac{1}{4} \cdot 4 = 1$
so this checks out.
Now we assume the theorem is true for some n = k. That is:
$\displaystyle \sum_{i = 1}^k i^3 = \frac{1}{4} \cdot k^2(k + 1)^2$
So we want to show this for n = k + 1.
$\displaystyle \sum_{i = 1}^{k + 1} i^3 = \sum_{i = 1}^k i^3 + (k + 1)^3 = \frac{1}{4} \cdot k^2(k + 1)^2 + (k + 1)^3$
You need to show that this is the same thing as:
$\displaystyle \frac{1}{4} \cdot (k + 1)^2((k + 1) + 1)^2 = \frac{1}{4} \cdot (k + 1)^2(k + 2)^2$
-Dan