hryggcx

2022-07-11

Sum of cubed roots
${x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}$
${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}$
where ${x}_{1},{x}_{2},{x}_{3}$ are the roots of
${x}^{3}+2{x}^{2}+3x+4=0$
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
$\left({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}\right)\left({x}_{1}+{x}_{2}+{x}_{3}\right)$
but that did work.

Dobermann82

Expert

If ${x}_{1},{x}_{2},{x}_{3}$ are the roots of ${x}^{3}+2{x}^{2}+3x+4=0$ then
${x}^{3}+2{x}^{2}+3x+4=\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)\left(x-{x}_{3}\right)$
$={x}^{3}-\left({x}_{1}+{x}_{2}+{x}_{3}\right){x}^{2}+\left({x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3}\right)x-{x}_{1}{x}_{2}{x}_{3}={x}^{3}-{e}_{1}{x}^{2}+{e}_{2}x-{e}_{3}.$
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 $\left\{{x}_{1}+{x}_{2}+{x}_{3},{x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3},{x}_{1}{x}_{2}{x}_{3}\right\}$
In the case of the fourth power sums you should get
${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}={e}_{1}^{4}-4{e}_{1}^{2}{e}_{2}+4{e}_{1}{e}_{3}+2{e}_{2}^{2}=18$

Esmeralda Lane

Expert

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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