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I have the following nonlinear system:

$$x+y+z=1$$ $$x^2+y^2+z^2=2$$ $$x^3+y^3+z^3=3$$ Find $x^5+y^5+z^5$.

So it was basically a long string of raising each equation to a higher power, and using substitution to reduce down each remaining expansion. For example, $(x+y+z)^2=1$ and from here i found that

$$xy+xz+yz=-\frac{1}{2}$$

Then $(x+y+z)^3=1$ implied that

$$xy^2+zx^2+x^2y+x^2z+y^2z+yz^2+2xyz=-\frac{2}{3}$$

Then noting that $(x^2+y^2+^2)(x^3+y^3+z^3)=6$, expanding out would give me a number of terms involving 5 degrees. The work it took was a couple of pages long, but I was able to show that $x^5+y^5+z^5=x^4+y^4+z^4$. Then i used $(x^2+y^2+z^2)^2=4$ to show ultimately that.

$$x^5+y^5+z^5=4-2(x^2y^2+x^2z^2+y^2z^2)$$

Closing in, I then showed that if $(xy+xz+yz)^2=\frac{1}{4}$, then

$$x^2y^2+x^2z^2+y^2z^2=\frac{1}{4}-2xyz$$

Thus,

$$x^5+y^5+z^5=4-2\left(\frac{1}{4}-2xyz\right)=\frac{7}{2}+2(2xyz)$$

Finally, from above, I could solve my second equation for $2xyz$ and substitute that in giving

$$x^5+y^5+z^5=\frac{7}{2}-\frac{4}{3}-2(xy^2+xz^2+x^2y+xz^2+y^2z-yz^2)$$

$$x^5+y^5+z^5=\frac{13}{6}-2[x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)]$$

$$x^5+y^5+z^5=\frac{13}{6}-2[x(2-x^2)+y(2-y^2)+z(2-z^2)]$$

$$x^5+y^5+z^5=\frac{13}{6}-2[2(x+y+z)-(x^3+y^3+z^3)]$$

$$x^5+y^5+z^5=\frac{13}{6}-2(2-3)$$

And so I believe that

$$x^5+y^5+z^5=\frac{25}{6}$$

Is this correct? Would there be any simpler way (although for what it's worth, it wasn't 3 full pages) without direct calculation and substitution?

Mando
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Assume that the solutions of the equation $X^3 + pX^2 + qX + r = 0$ are $x, y$ and $z$. See $f(X) = X^3 + pX^2 + qX + r$ as a polynomial. By Vieta's Formula, $p = -(x+y+z)$ and $q = xy + yz + zx$, which you already know. Substitute $x, y$, and $z$ to the equation. Add the then-obtained equations, and use the given equations to solve for the values of $r$. Now, find $x^4+y^4+z^4$, and subsequently $x^5+y^5+z^5$, by multiplying $X$ and $X^2$ to the equation $X^3 + pX^2 + qX + r = 0$.

Shubhav Jain
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  • But after substitution and summation I only get one equation in $p,q$, and $r$. $$3+2p+3q+3r=0$$ And I can't solve this exclusively. So what am I missing from your explanation? – Mando Nov 06 '23 at 15:33
  • I have made some edits to the answer. Check it out. – Shubhav Jain Nov 06 '23 at 15:41
  • Okay. Vieta. I get it. – Mando Nov 06 '23 at 15:44
  • And I found my dumb error.... the downside of working with MY method is that there are many places to make an error in just directly calculating. – Mando Nov 06 '23 at 15:59