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I need to calculate the sums

$$x_1^3 + x_2^3 + x_3^3$$


$$x_1^4 + x_2^4 + x_3^4$$

where $x_1, x_2, x_3$ are the roots of


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


but that did work.

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up vote 12 down vote accepted

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 $x_1^4 + x_2^4 + x_3^4 = e_1^4 - 4 e_1^2 e_2 + 4 e_1 e_3 + 2 e_2^2 = 18$.

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I've read the details but there is no explanation how you reached the end result, I'm more interested in that – andrei Apr 2 '11 at 13:12
solved it.the trick is that after you solve x1^2+x2^2+x3^2 then u procede to write (x1^2+x2^2+x3^2)*(x1+x2+x3) and get x1,2,3^3 from that equation.(you'll end up having some terms like x1^2x2 and x1^2x3 do not replace them with the first equation the sum of the roots, use the second equation and third) – andrei Apr 2 '11 at 14:37
Fluent and concise answer, +1!! – awllower Jan 2 '12 at 14:40
+1 very nice and interesting that your last line was also part of one of my questions, despite I had $4$ variables and an additional $-4e_4$. Why is this missing in your case? Ah, I see $x_4=0$. Still nice. – draks ... Apr 8 '12 at 18:34

I think what you need is Newton's identities, in particular the section about their application to the roots of a polynomial.

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