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$x+y+z=0$

$x^3+y^3+z^3=9$

$x^5+y^5+z^5=30$

$xy+yz+zx=?$

I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?

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    @MichaelHardy I see. The question has already quite the upvotes. Seems many agree with you!2012-04-29

2 Answers 2

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We have the Newton-Girard identities $x^3+y^3+z^3=(x+y+z)^3+3xyz-3(x+y+z)(xy+xz+yz)$ and $\begin{split}x^5+y^5+z^5=&(x+y+z)^5-5(x+y+z)^3 (xy+xz+yz)+\\5(x+y+z)&(xy+xz+yz)^2-5xyz(xy+xz+yz)+5xyz(x+y+z)^2\end{split}$ Replacing all instances of $x+y+z$ with $0$, we have the simultaneous equations

$\begin{align*} 3xyz&=9\\ -5xyz(xy+xz+yz)&=30 \end{align*}$

You should now be able to solve for what you need.

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    @MarkBennet Thanks, I got it. It was derived from the similar way!2012-04-29
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Here is one way of making progress, which uses the cubic as part of the solution. There are other routes which involve knowing some standard factorisations.

First note that $z=-(x+y)$ from the first equation and substitute in the second, obtaining:

$-3x^2y-3xy^2 = 9$

Divide by 3 to get:

$-xy(x+y) = xyz = 3$

Now $x,y,z$ are the roots of the cubic equation $t^3+kt-3 = 0$,

and therefore satisfy $t^5+kt^3-3t^2=0$

Substitute $x,y,z$ successively into this equation and add to get

$30+9k-3(x^2+y^2+z^2) = 0$

And use $0=(x+y+z)^2=x^2+y^2+z^2+2k$ to finish.

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    The advantage is that it keeps the manipulations simple and reduces the opportunity for error. So it is a practical method for simple cases.2012-04-28