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.
 A: Use Viete formulas: 
$$\alpha\beta = c/a$$$$\alpha + \beta = - b/a$$
Therefore $$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha^2\beta - 3\alpha\beta^2 = (-b/a)^3 + 3bc/a^2$$
A: First note that $\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$ and also note that $-\frac{b}{a}=\alpha+\beta$ and $\frac{c}{a}=\alpha\beta$ (do you see why?) We can make $$\alpha^2+2\alpha\beta+\beta^2=(\alpha+\beta)^2=\frac{b^2}{a^2}$$ so our final outcome will be \begin{align}
\alpha^3+\beta^3&=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)\\
&= -\frac{b}{a}(\alpha^2+2\alpha\beta+\beta^2-3\alpha\beta)\\
&= -\frac{b}{a}(\frac{b^2}{a^2}-3\frac{c}{a})\\
&= -\frac{b^3-3abc}{a^3}
\end{align}
A: For $\frac1{\alpha^3}+\frac1{\beta^3}$, use that the roots of $a+bx+cx^2$ are $\frac1\alpha$ and $\frac1\beta$ to reduce to the previous problem.
A: 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
$\alpha^3 + \beta^3 = 
   \dfrac{(b^2 - ac) \left(-\dfrac ba \right) + 2bc}{a^2} =
   \dfrac{(-b^3 + abc) + 2abc}{a^3} =
   \dfrac{3abc - b^3}{a^3}$
