[Math] The sum of the cubes of the reciprocal values of the roots of the equation $x^2+ax+1=0$ is

Question

The equation :
$$x^2+ax+1=0$$

$a$ is a real number.

How to find the sum of cubes of the reciprocal of roots for this equation.

I tried solving this just by brute forcing it, but I get expressions that are ugly and pretty sure would yield nothing in the long run. So there is probably a trick to doing this, as the possible solutions I have nice expressions.

Any hints would also be fine.

Answer 1

Hint: Let $\alpha$ and $\beta$ be the roots. Then $${1\over \alpha ^3}+{1\over \beta ^3}={\alpha^3+\beta^3 \over{\alpha^3\beta^3}} ={(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \over{(\alpha\beta})^3} $$