What is the sum of fifth power of roots of the equation $$x^3+qx+r$$.I tried expanding $$(a+b+c)^5$$ but it didn't work instead it is becoming more and more complex.


$$x^3+qx+r=0\implies a+b+c=0, ab+bc+ca=q,abc=-r$$

$$\implies \sum a^2=(a+b+c)^2-2(ab+bc+ca)$$

and multiplying the given eqaution by $x^n$ $$ x^{n+3}+qx^{n+1}+rx^n=0$$

$$\implies a^{n+3}+qa^{n+1}+ra^n=0$$

$$\implies\sum a^{n+3}+q\sum a^{n+1}+r\sum a^n=0$$

$$n=0\implies\sum a^3+q\sum a+r\sum 1=0\iff \sum a^3=-q\cdot0-r\cdot3$$

$$n=2\implies\sum a^5+q\sum a^3+r\sum a^2=0\iff \sum a^5=-q\sum a^3-r\sum a^2$$

Reference : Newton's Sums

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By Viete's relations,




Since roots satisfy the equation, we have, $$a^3+qa+r=0$$

$$b^3+qb+r=0$$ $$c^3+qc+r=0$$

Or equivalently,


If we denote $S_n=a^n+b^n+c^n$,adding these three equations, we have,


$$S_5=-qS_3-rS_2 = -q(-qS_1-rS_0)-rS_2$$

We have $S_1=0$ and $S_2=(a+b+c)^2-2(ab+bc+ca)=-2q$


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  • $\begingroup$ Why so long? If we already know $\;a+b+c=0\;$ , then trivially $\;(a+b+c)^5=0\;$ ...! Am I missing something trivial here? $\endgroup$ – DonAntonio Mar 24 '14 at 18:14
  • $\begingroup$ @DonAntonio $a^5+b^5+c^5$ not $(a+b+c)^5$ Btw it isn't long. almost zero computation. I just spaced out the formatting. $\endgroup$ – Guy Mar 24 '14 at 18:15
  • $\begingroup$ Oh, I see...yet I went by what the OP wrote in his question. Thanks. $\endgroup$ – DonAntonio Mar 24 '14 at 18:18

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