Find $x_1^3+x_2^3+x_3^3$ for a degree 3 polynomial.

Question

I have the polynomial $P(x) = x^3+mx^2-3x+1, m\in \mathbb{R}$. I need to find $x_1^3+x_2^3+x_3^3$ as a $m$ function.

I tried to use Viette equations: $x_1+x_2+x_3 = m, x_1x_2+x_1x_3+x_2x_3 = 3, x_1x_2x_3 = 1$ Then I expanded $(x_1+x_2+x_3)^3 = (x_1^3+x_2^3+x_3^3) + 3(x_1^2x_2+x_1^2x_3+x_1x_2^2+x_2^2x_3+x_1x_3^2+x_2x_3^2) + 6x_1x_2x_3$. After that I expanded $(x_1x_2+x_1x_3+x_2x_3)^2 = 2(x_1x_2x_3)+x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2$. I don't know if my way of doing it is the right way, and if it is can you help from this point on?

Answer 1

We have that $x^3_1+x^3_2+x^3_3$ is symmetric (fixed by any element of $S_3$).

Therefore, you may express $x^3_1+x^3_2+x^3_3$ using only the elementary symmetric polynomials.

You may use symmetric reduction on wolframalpha to deduce that:

$x^3_1+x^3_2+x^3_3=(x_1+x_2+x_3)^3-3(x_1x_2+x_2x_3+x_1x_3)(x_1+x_2+x_3)+3x_1x_2x_3$.

Now plug in to get $x^3_1+x^3_2+x^3_3$ in terms of the coefficients.

I believe you should get $(-m)^3-3(-3)(-m)+3(-1)=-m^3-9m-3$

Answer 2

Since $x_1,x_2,x_3$ are the roots we find that $x_1^3=3x_1-mx_1^2-1$ similarly for $x_2,x_3$ adding the three equations we find that $x_1^3+x_2^3+x_3^3=3(x_1+x_2+x_3)-m(x_1^2+x_2^2+x_3^2)-3$ we see that $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_1x_3)$ thus the required value is $x_1^3+x_2^3+x_3^3=3(-m)-m(((-m)^2)-2(-3))-3=-3m-m^3-6m-3=-m^3-9m-3$