hryggcx

Answered

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

$({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2})({x}_{1}+{x}_{2}+{x}_{3})$

but that did work.

${x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}$

${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}$

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

$({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2})({x}_{1}+{x}_{2}+{x}_{3})$

but that did work.

Answer & Explanation

Dobermann82

Expert

2022-07-12Added 15 answers

If ${x}_{1},{x}_{2},{x}_{3}$ are the roots of ${x}^{3}+2{x}^{2}+3x+4=0$ then

${x}^{3}+2{x}^{2}+3x+4=(x-{x}_{1})(x-{x}_{2})(x-{x}_{3})$

$={x}^{3}-({x}_{1}+{x}_{2}+{x}_{3}){x}^{2}+({x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3})x-{x}_{1}{x}_{2}{x}_{3}={x}^{3}-{e}_{1}{x}^{2}+{e}_{2}x-{e}_{3}.$

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 $\{{x}_{1}+{x}_{2}+{x}_{3},{x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3},{x}_{1}{x}_{2}{x}_{3}\}$

In the case of the fourth power sums you should get

${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}={e}_{1}^{4}-4{e}_{1}^{2}{e}_{2}+4{e}_{1}{e}_{3}+2{e}_{2}^{2}=18$

${x}^{3}+2{x}^{2}+3x+4=(x-{x}_{1})(x-{x}_{2})(x-{x}_{3})$

$={x}^{3}-({x}_{1}+{x}_{2}+{x}_{3}){x}^{2}+({x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3})x-{x}_{1}{x}_{2}{x}_{3}={x}^{3}-{e}_{1}{x}^{2}+{e}_{2}x-{e}_{3}.$

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 $\{{x}_{1}+{x}_{2}+{x}_{3},{x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{2}{x}_{3},{x}_{1}{x}_{2}{x}_{3}\}$

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

2022-07-13Added 7 answers

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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