<s>[INST] How to show $\frac{1}{\omega-3\omega^2}=\frac{4\omega^2+3}{13}$. I was solving a cubic equation with two different methods and one of the root I got with first method is $\frac{1}{\omega-3\omega^2}$ and with second method I got $\frac{4\omega^2+3}{13}$.
$\omega$ is the cube root of unity
If I set one equals another then I got an equality $13=13$ which indicates that they are equal.
$$\frac{1}{\omega-3\omega^2}=\frac{\omega^2}{1-3\omega}$$
After this I think we have to do rationalization kind of thing to the make denominator real but I could not figure out How.
Thank you for your help!
 [/INST] The conjugate of $\omega$ is $\omega^2$, so multiply top and bottom by $\omega^2-3\omega$. This gives
$$\frac{1}{\omega-3\omega^2}$$
$$=\frac{1}{\omega-3\omega^2}\cdot\frac{\omega^2-3\omega}{\omega^2-3\omega}$$
$$=\frac{\omega^2-3\omega}{1-3\omega^2-3\omega^4+9}$$
Use $\omega^3=1$ and $1+\omega+\omega^2=0$ to give
$$=\frac{\omega^2-3\omega}{1-3(\omega^2+\omega)+9}$$
$$=\frac{\omega^2-3\omega}{13}$$
$-3\omega=3+3\omega^2$ then gives your result.
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