Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation. I have a question.
Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$
I don't know how to proceed. Any help would be appreciated.

Just to be different.
If $\alpha$ is a solution of $ax^2 + bx + c = 0$ Then
$a\alpha^2 + b\alpha + c = 0$
So
$\quad \alpha^2 = -\dfrac{b\alpha + c}{a}$
$\quad \alpha^3 = -\dfrac{b\alpha^2 + c\alpha}{a}
                = -\dfrac{b\left( -\dfrac{b\alpha + c}{a} \right)+ c\alpha}{a}
                = -\dfrac{ -b^2\alpha - bc + ac\alpha}{a^2}   
                = -\dfrac{(ac -b^2)\alpha - bc}{a^2}$ 
Similarly,
$\quad \beta^3 = -\dfrac{(ac -b^2)\beta - bc}{a^2}$ 
So $\alpha^3 + \beta^3 = 
   -\dfrac{(ac -b^2)(\alpha + \beta) - 2bc}{a^2} =
   \dfrac{(b^2 - ac)(\alpha + \beta) + 2bc}{a^2}$
Since $\alpha + \beta = -\dfrac ba$, we see that
$\alpha^3 + \beta^3 = 
   \dfrac{(b^2 - ac) \left(-\dfrac ba \right) + 2bc}{a^2} =
   \dfrac{(-b^3 + abc) + 2abc}{a^3} =
   \dfrac{3abc - b^3}{a^3}$
