$$\sum\limits_{i=1}^n i^3 = \left(\frac{n(n + 1)}{2}\right)^2$$
My basis step is $P(1)$ sets the $LHS = RHS = 1$.
For the inductive step, I assume $n = k$ holds for $k+1$. On the $RHS$:
$$\left(\frac{(k + 1)((k + 1) + 1)}{2}\right)^2$$
But I don't know how to convert the summation into something that can evaluated algebraically.
Disclaimer: this is a question from an exam review sheet.
Base case: $n=1$,
$$\sum_{i=1}^1i^3=1^3=1=1^2=\left(\frac{2}{2}\right)^2=\left(\frac{1(1+1)}{2}\right)^2.$$
Induction: Suppose it holds for some $(n-1)\geq 1$. Then,
$$\sum_{i=1}^{n}i^3=n^3+\sum_{i=1}^{n-1}i^3 =n^3+\left(\frac{(n-1)n}{2}\right)^2=n^3+\frac{n^4-2n^3+n^2}{4}$$
$$=\frac{n^4+2n^3+n^2}{4}=\left(\frac{n(n+1)}{2}\right)^2.$$
Note: I did the induction by showing that if the relationship holds for $n-1$ then it does as well for $n$ instead of the usual 'if it holds $n$ then it holds for $n+1$' purely out of convenience (to avoid expanding $(n+1)^3$). However, if you want, you can show that 'if it holds $n$ then it holds for $n+1$' pretty much in an identical fashion.
I assume that $P(n)$ means that the formula holds for $n$.
You assume that this holds for $n=k$ and you want to show that this holds for $n = k+1$. On the right hand side you indeed have what you have written.
One the left hand side you have $$ \sum_{i=1}^{k+1} i^3 = \left[\sum_{i=1}^{k} i^3\right] + (k+1)^3 $$ Now you can use the induction hypothesis and continue to get $$ \left[\sum_{i=1}^{k} i^3\right] + (k+1)^3 = \left(\frac{k(n+1)}{2}\right)^2 + (k+1)^3. $$ All that is left for you to show is that $$ \left(\frac{k(n+1)}{2}\right)^2 + (k+1)^3 $$ is equal to the right hand side that you have in your question.
