<s>[INST] 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.
 [/INST] Every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials, in this case $s_1=x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$ and $s_3=x_1x_2x_3$. Since $x_1^3+x_2^3+x^3$ is homogeneous, we can find $a$, $b$ and $c$ such that
$$
x_1^3+x_2^3+x_3^3=as_1^3+bs_1s_2+cs_3
$$


*

*For $x_1=1$, $x_2=0$, $x_3=0$: $1=a$

*For $x_1=1$, $x_2=1$, $x_3=0$: $2=8a+2b$

*For $x_1=1$, $x_2=1$, $x_3=1$: $3=27a+9b+c$


Therefore $a=1$, $b=-3$, $c=3$ and finally
$$
x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3
$$
This is a general result.
In your case, by Viète’s formulas
$$
x_1+x_2+x_3=0,\qquad
$$
so in the end
$$
x_1^3+x_2^3+x_3^3=3x_1x_2x_3
$$
</s>