Maclaurin expansion of a rational function

Question

I have to find the Maclaurin expansion of $$g(x)=\frac{x^{2}}{x^{6}+2x^{3}+1}$$ I know $$x^{6}+2x^{3}+1=(1+x^{3})^{2}$$ then $$g(x)=\left(\frac{x}{1+x^{3}}\right)^{2}$$ But I have no idea how to find it. I tried partial fractions, but it seemed it'd make harder to do.

Answer 1

As said in comments $$g(x)=\left(\frac{x}{1+x^{3}}\right)^{2}=-\frac 13 \left(\frac{1}{1+x^{3}}\right)'$$ For the time being, let $y=x^3$ and remember that $$\frac{1}{1+y}=\sum_{n=0}^\infty (-1)^n y^n\implies \frac{1}{1+x^{3}}=\sum_{n=0}^\infty (-1)^n x^{3n}$$ Then $$g(x)=-\frac 13\left(\sum_{n=0}^\infty (-1)^n x^{3n} \right)'=-\frac 13\sum_{n=0}^\infty (-1)^n 3n \,x^{3n-1}=\sum_{n=0}^\infty (-1)^{n-1} n\, x^{3n-1}$$