Sum of cubed roots I need to calculate the sums
$$x_1^3 + x_2^3 + x_3^3$$
and   
$$x_1^4 + x_2^4 + x_3^4$$
where $x_1, x_2, x_3$ are the roots of 
$$x^3+2x^2+3x+4=0$$
using Viete's formulas.
I know that $x_1^2+x_2^2+x_3^2 = -2$, as I already calculated that, but I can't seem to get the cube of the roots. I've tried
$$(x_1^2+x_2^2+x_3^2)(x_1+x_2+x_3)$$ 
but that did work.

If $x_1,x_2,x_3$ are the roots of $x^3+2x^2+3x+4=0$ then $$x^3+2x^2+3x+4 = (x-x_1)(x-x_2)(x-x_3) $$ $$= x^3 - (x_1 + x_2 + x_3)x^2 + (x_1 x_2 + x_1 x_3 + x_2 x_3)x - x_1 x_2 x_3 = x^3 - e_1 x^2 + e_2 x - e_3.$$ So $e_1 = -2$, $e_2 = 3$ and $e_3 = -4$.
Now the trick is to express the power sums $x_1^3 + x_2^3 = x_3^3$ and $x_1^4 + x_2^4 = x_3^4$ in terms of the elementary symmetric polynomials $\{x_1 + x_2 + x_3,x_1 x_2 + x_1 x_3 + x_2 x_3,x_1 x_2 x_3\}$.
See my answer to the question here for details on how to do that Three-variable system of simultaneous equations
In the case of the fourth power sums you should get $x_1^4 + x_2^4 + x_3^4 = e_1^4 - 4 e_1^2 e_2 + 4 e_1 e_3 + 2 e_2^2 = 18$.
