$x^3+x+m=0$, $x_1^5+x_2^5+x_3^5=10$, $m=?$

Question

Given the equation:

$x^3+x+m=0$, $m \in R$

$x_1,$ $x_2$ and $x_3$ are roots of the equation,

$x_1^5+x_2^5+x_3^5=10.$

Find a value of $m$.

I've applied Vieta's a couple of times and got:

$x_1^4+x_2^4+x_3^4=2$

$x_1^4x_2^4+x_1^4x_3^4+x_2^4x_3^4=1+4m^2$

$x_1^4x_2^4x_3^4=m^4$

But kind of stopped here, as I don't know how I am supposed to find the sum of odd powers...

Could I have some hints on how to do this? Preferably using the method I tried if it can be done this way, but others will be just as helpful. Thank you.

Answer 1

By the Viete's theorem we obtain: $$x_1+x_2+x_3=0,$$ $$x_1x_2+x_1x_3+x_2x_3=1$$ and $$x_1x_2x_3=-m.$$ Thus, $$x_1^5+x_2^5+x_3^5=-5(x_1x_2+x_1x_3+x_2x_3)x_1x_2x_3=5m,$$ which gives $m=2$.

I used the following statement.

Let $x_1+x_2+x_3=p$, $x_1x_2+x_1x_3+x_2x_3=q$ and $x_1x_2x_3=r$.

Thus, $$x_1^5+x_2^5+x_3^5=p^5-5p^3q+5pq^2+5u^2r-5qr.$$

Answer 2

$x_i^5=-x_i^3-mx_i^2=x_i+m-mx_i^2.$ As $x_1+x_2+x_3=0$ and $x_1^2+x_2^2+x_3^2=-2$, then $$x_1^5+x_2^5+x_3^5=3m+2m=5m$$ so $m=2$.

Answer 3

$$\begin{align} x^3+x+m=0 &\implies x^5+x^3+mx^2=0\\ &\implies x^5-(x+m)+mx^2=0\\ &\implies m(x^2-1)=x-x^5\\ &\implies m(x_1^2+x_2^2+x_3^2-3)=(x_1+x_2+x_3)-(x_1^5+x_2^5+x_3^5) \end{align}$$

Now $x_1+x_2+x_3=0$ and

$$x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_2x_3+x_3x_1)=0^2-2\cdot1=-2$$

by Vieta, and $x_1^5+x_2^5+x_3^5=10$ by assumption, so

$$m(-2-3)=0-10$$

and thus $m=2$.