I need to solve the recurrence:
$$ \begin{align*} T(n) &= kT\left(\frac{n}{2}\right) + (k - 2)n^3 \\ &\textit{where}\; k \in \mathbb{Z}: k \geq 2 \\ &= k(kT\left(\frac{n}{4} + \frac{(k - 2)n^3}{2}\right) + (k - 2)n^3 \\ &= k^2T\left(\frac{n}{4}\right) + \frac{(k + 2)(k - 2)n^3}{2} \\ &= k^2\left(kT\left(\frac{n}{8}\right) + \frac{(k + 2)(k - 2)n^3}{8}\right) + \frac{(k + 2)(k - 2)n^3}{2} \\ &= k^3T\left(\frac{n}{8}\right) + \frac{(k^2 + 4)(k + 2)(k - 2)n^3}{8} \\ &\dots \\ &= k^iT\left(\frac{n}{2^i}\right) + \frac{(k - 2)n^3}{2^i}\prod_{j=1}^i(k^j+2^j) \end{align*} $$
But I don't know how can I develop this further. I know that the recursion will end at $i = \lg n$, but I don't know what to do with the product term.
Here is a quick way to bound many kinds of functions defined by a recurrence if one is interested in the function's behaviour for large $n$. Let $\lg x=\log_2 x$. Suppose we are given a recurrence of the form $f(n)=cf(n/2)+p(n), f(1)=1$, where $p(n)$ is a function depending on $n$ and $c$ is a constant. If $p(n)\ge 0$ for all $n\ge 1$, then $f(n)\ge cf(n/2)$, so $f(n)\ge c^{\lg n}= n^{\lg c}$. Further, if $p(n)\ge p(n/2)$ for every $n\ge 2$, then $f(n)\le c^{\lg n} + p(n)(c^{\lg n}-1)/(c-1)\le n^{\lg c}(1+p(n)/(c-1))$. So $f(n)=\Omega(n^{\lg c})$ and $f(n)=O(n^{\lg c}p(n))$, and it is easy to see that these expressions also hold for other constant values of $f(1)>0$.
The specific recurrence has $p(n)=(k-2)n^3$ and $c=k\ge 2$, which satisfy the conditions assumed for the inequalities to hold. In particular, $f(n)=O(n^{3+\lg k})$ and $f(n)=\Omega(n^{\lg k})$ which is a polynomial in $n$ for fixed $k$.
(As stated the sequence of equations in the question has errors, and the final expression is $\Omega(n^{\lg n})$ which is not polynomial in $n$.)