The question is to find a polynomial expression in $p$ for the sequence $$q_p = \sum_{\ell=0}^p \ell^3$$ from the geometric series $\sum_p x^p$.
I tried induction but apparently there's another way to do it via convolution and I have no idea how
I know the answer is p^2(p+1)^2 all over 4, I just got that off a generic internet formula for the sum of cube series, but I have no idea how to derive it in this question.
This might not be exactly what your textbook authors had in mind, but it’s a nice exercise in the use of generating functions. The idea is that you treat a sequence as the coefficients of the power series. The function that corresponds to this power series in a sense encodes the sequence. (There are other types of generating functions, but for this problem I’ll only focus on ordinary power series.) Many common operations on sequences correspond to operations on their generating functions, which are often much easier to manipulate than the original sequences. For example, convolution of sequences corresponds to multiplication of their generating functions.
In particular, the generating function of the infinite sequence that consists entirely of $1$s is $\sum_{k=0}^\infty x^k = \frac1{1-x}$. The convolution of any sequence with this one produces the sequence of partial sums of the original sequence, therefore if $f(x)$ is the generating function of a sequence, $\frac1{1-x}f(x)$ is the generating function of its partial sums. Differentiating $ax^n$ gives $nax^{n-1}$ and multiplying a generating function by $x$ shifts the sequence to the right, so $xD$ is the “multiply by $n$” operator. Putting all of this together, we can construct the generating function for the sequence of the sums of the first $n$ cubes: $$f(x) = \frac1{1-x}(Dx)^3\frac1{1-x} = {x(1+4x+x^2)\over(1-x)^5}.$$ Working from left to right, we start with the sequence of all $1$s, apply the “multiply by $n$” operator three times to get a sequence of cubes, then multiply by $1/(1-x)$ to form its partial sums. Producing the expression on the right from this is a straightforward matter of applying the product and quotient rules for derivatives a few times. So, the sum of the first $n$ cubes is the coefficient of $x^n$ in the power series of $f$, usually denoted by $[x^n]\,f(x)$.
Working out a general formula for the coefficients of the power series for this function looks tedious and doesn’t seem like any less work than trying to come up with a formula for the first $n$ cubes directly, but there’s a handy identity involving binomial coefficients that simplifies the task immensely: $$[y^n]{y^k\over(1-y)^{k+1}}=\binom nk.$$ Using this identity, we have $$[x^n]{1\over(1-x)^5} = \binom{n+4}4$$ and so $$\sum_{k=0}^n n^3 = [x^n]\,f(x) = \binom{n+3}4+4\binom{n+2}4+\binom{n+1}4 = {n^2(n+1)^2\over4}.$$