Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x_3) = x^3+px+q = 0$$
Now find the coefficient near the powers of $x$:
$$
x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 = x^3+px+q
$$
That means that I can write a system of equations:
$$
\begin{cases}
-(x_1 + x_2 + x_3) = 0 \\
x_1x_2 + x_1x_3 + x_2x_3 =  p \\
- x_1x_2x_3 = q
\end{cases}
$$
At this point I got stuck. I've tried to raise $x_1 + x_2 + x_3$ to 3 power and expand the terms, but that didn't give me any insights. It feels like I have to play with the system of equations in some way but not sure what exact.

Rewrite $x_1,x_2,x_3$ with $a,b,c$. From first Vieta formula we have $$a+b+c=0$$
so $a+b=-c$ and so on...
Now $$a^3+b^3+c^3= (a+b)(a^2-ab+b^2)+c^3 = c(\underbrace{-a^2+ab-b^2+c^2}_I)$$
Since
$$I = -a^2+ab-b^2+c^2 = a(b-a)+(c-b)(c+b) = $$ $$a(b-a)-a(c-b) = a(2b-a-c)=a(2b+b)=3ab$$ 
