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So I have this equation : $2x^3-7x^2+5x-6=0$ and I have to calculate $x_1^4+x_2^4+x_3^4$ but I don't know the formula of this one. I know that $S_1 = x_1+x_2+x_3 = -b/a$ $S_2=x_1*x_2+x_1*x_3+x_2*x_3 = c/a$

$S_3 = x_1*x_2*x_3 = -d/a$

and

$x_1^3+x_2^3+x_3^3 = S_1^3 - 3S_1S_2 + 3S_3$

but what about $x_1^4+x_2^4+x_3^4 = ?$

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  • $\begingroup$ Did you notice that there is an obvious root for the equation ? $\endgroup$ – Claude Leibovici Jun 21 '15 at 15:37
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    $\begingroup$ Check en.m.wikipedia.org/wiki/Newton's_identities $\endgroup$ – Macavity Jun 21 '15 at 15:42
  • $\begingroup$ @ClaudeLeibovici I don't need to find the root of the equation, I just need to compute $x_1^4+x_2^4+x_3^4$ with the help of Viete equations $\endgroup$ – southpaw93 Jun 21 '15 at 15:43
  • $\begingroup$ the result should be $\frac{1313}{16}$ $\endgroup$ – Dr. Sonnhard Graubner Jun 21 '15 at 15:44
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    $\begingroup$ Well you can try learning the method, or looking at the formula m.wolframalpha.com/input/… Your choice. $\endgroup$ – Macavity Jun 21 '15 at 16:36
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we have $$3^4+(1/4+1/4\sqrt{15}i)^4+(1/4-1/4\sqrt{15}i)^4=\frac{1313}{16}$$

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  • $\begingroup$ it is true, give me your solution $\endgroup$ – Dr. Sonnhard Graubner Jun 21 '15 at 17:32
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    $\begingroup$ @CioroianuDenis that should be $s_1^4-4s_1^2s_2+...$ $\endgroup$ – Macavity Jun 21 '15 at 18:05
  • $\begingroup$ @CioroianuDenis now you have to check your arithmetic. $\endgroup$ – Macavity Jun 21 '15 at 18:15
  • $\begingroup$ ok i did not downvoted $\endgroup$ – Dr. Sonnhard Graubner Jun 21 '15 at 18:25

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