Simplify the quotient of powers $\frac{x^5\left(x-3\right)^8}{\left(x^2+3\right)^7}$

We can expand the expression $\left(x-3\right)^8$ using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer $n$. The formula is as follows: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$. The number of terms resulting from the expansion always equals $n + 1$. The coefficients $\left(\begin{matrix}n\\k\end{matrix}\right)$ are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of $a$ decreases, from $n$ to $0$, while the exponent of $b$ increases, from $0$ to $n$. If one of the binomial terms is negative, the positive and negative signs alternate.