<s>[INST] How to split $\frac{x^3}{(x^2+5)^2}$ into partial fractions? 
Split $\frac{x^3}{(x^2+5)^2}$ into partial fractions.

I have tried letting $$\frac{x^3}{(x^2+5)^2}=\frac{ax+b}{x^2+5}+\frac{cx^3+dx^2+ex+f}{(x^2+5)^2},$$ but this yields a system of only four simultaneous equations. 
I have also tried looking at $$\frac{x^3}{(x^2+5)^2} = x \cdot \left( \frac{x}{x^2+5} \right) ^2,$$ but sadly in vain.
 [/INST] If we format it into the partial fractions 'template':
$$\frac{x^3}{(x^2+5)^2}=\frac{Ax+B}{(x^2+5)}+\frac{Cx+D}{(x^2+5)^2}$$
Multiplying both sides by the denominator, 
$$\frac{x^3\left(x^2+5\right)^2}{\left(x^2+5\right)^2}=\frac{\left(B+Ax\right)\left(x^2+5\right)^2}{x^2+5}+\frac{\left(D+Cx\right)\left(x^2+5\right)^2}{\left(x^2+5\right)^2}$$
Simplifying, we find that:
$$x^3=\left(B+Ax\right)\left(x^2+5\right)+Cx+D$$
so
$$x^3=Ax^3+Bx^2+5Ax+Cx+5B+D$$
$$x^3=Ax^3+Bx^2+x\left(5A+C\right)+\left(5B+D\right)$$
We have the system of equations
$$\begin{bmatrix}5B+D=0\\ 5AC=0\\ B=0\\ A=1\end{bmatrix}$$
Solving this gives $D=0,\:C=-5,\:B=0,\:A=1$. 
Putting this in our original 'template' gives us:
$$\frac{x}{x^2+5}-\frac{5x}{\left(x^2+5\right)^2}$$
</s>