The sum of the cubes of the reciprocal values of the roots of the equation $x^2+ax+1=0$ is? 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.
 A: Hint: Let $\alpha$ and $\beta$ be the roots. Then 
$${1\over \alpha ^3}+{1\over \beta ^3}={\alpha^3+\beta^3 \over{\alpha^3\beta^3}} ={(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \over{(\alpha\beta})^3} $$
A: According to Viete's formulas
$x_1+x_2=-a$
$x_1x_2=1$
We are looking for 
$\frac{1}{x_1^3}+\frac{1}{x_2^3}$
But because $x_1x_2=1$:
$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_2^3+x_1^3}{x_1^3 x_2^3}=x_2^3+x_1^3=(x_1+x_2)^3-3x1x_2^2-3x1^2x_2=(x_1+x_2)^3-3x_1-3x_2$
Now we substitute $x_1+x_2=-a$:
$(x_1+x_2)^3-3x_1-3x_2=-a^3+3a$
A: Brute forcing, though by no means the best way, is quite feasible here.  After all, this is just a quadratic equation.  The roots are $\frac{-a\pm\sqrt{a^2-4}}{2}$.
The reciprocals are $\frac{-a\mp\sqrt{a^2-4}}{2}$. This can be seen by flipping the roots over and rationalizing the denominator. Or more simply let $t=1/x$. Substitute in the equation. We get $t^2+at+1=0$.
Now we need to calculate
$$\left(  \frac{-a+\sqrt{a^2-4}}{2} \right)^3+\left(  \frac{-a-\sqrt{a^2-4}}{2} \right)^3.$$
Expand, using the ordinary expansion of $(s+t)^3$. There is pleasant cancellation, and we get
$$\frac{2}{8}\left(-a^3+3(-a)(a^2-4)       \right),$$
which simplifies to $3a-a^3$.
A: Let $y=\dfrac1{x^3}$
As $x^2+ax+1=0\iff x^2+ax=-1$
Cubing both sides,
$(x^2+ax)^3=(-1)^3\iff(x^3)^2+a^3(x^3)+3a(x^3)(-1)=-1$
$\implies\left(\dfrac1y\right)^2+(a^3-3a)\left(\dfrac1y\right)=-1$
$\implies y^2+(a^3-3a)y+1=0$
$\implies y_1+y_2=-\dfrac{a^3-3a}1$
