Relation between the roots and the coefficients of a polynomial I have studied that:
For the polynomial $ax^3+bx^2+cx+d=0$, with roots $\alpha, \beta, \gamma$:
We have:
$$\begin{align}
& \alpha + \beta + \gamma = -\frac ba \\
& \alpha\beta + \beta\gamma + \alpha\gamma = \frac ca \\ 
& \alpha\beta\gamma = -\frac da
\end{align}
$$
A question asks to prove for: $x^3-px-q=0$ that:
$$\begin{align}
& \alpha^2 + \beta^2 + \gamma^2 = 2p \\
& \alpha^3 + \beta^3 + \gamma^3 = 3q
\end{align}$$
I looked around the Internet but couldn't find a way to prove this. How would I go around doing this?
 A: $$\left(\sum\alpha\right)^2=\sum\alpha^2+2\sum\alpha\beta$$
$$\iff\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta$$

$$a\alpha^3=-b\alpha^2-c\alpha-d$$
$$\implies a\sum\alpha^3=-b\sum\alpha^2-c\sum\alpha-3d$$
See also : Newton's Sums
A: The squared sum ($0$) minus twice the sum of products ($-p)$ is the sum of squares, $0-2(-p)=2p$.
From the equation, the cube equals $p$ times the value plus $q$. By summing, $p\cdot(0)+3q=3q$.
A: Let $f(x)=x^3−px−q=0 $ and assume that $\alpha, \beta,\gamma$ are the roots of the polynomial $ f(x)$ so 
\begin{align}
\alpha +\beta +\gamma =0\\
\alpha\beta +\beta\gamma +\gamma\alpha =-p\\
\alpha\beta\gamma=q
\end{align}
Consider,
\begin{align}
(\alpha +\beta +\gamma)^2 &= & 0\\
\implies \alpha^2+\beta^2+\gamma^2+2(\alpha\beta +\beta\gamma +\gamma\alpha) & =& 0\\
\implies \alpha^2+\beta^2+\gamma^2=-2(\alpha\beta +\beta\gamma +\gamma\alpha)=2p.
\end{align}
Similarly, we know that  $$ \alpha +\beta +\gamma =0 \implies \alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma=3q. $$
(This can also be shown by the similar argument that we have done in the square. Just consider $(\alpha +\beta +\gamma)^3$ and expand it you will get the result that $\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma $)
