12
$\begingroup$

$x+y+z=0$

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

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

$xy+yz+zx=?$

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?

  • 0
    I would first guess that the variables have to be integers. Then the third equation makes me think of $2^5=32 \approx 30$ and permutations of $(-1,-1,2)$ look interesting. Unfortunately they don't fit the cubic.2012-04-28
  • 0
    From $(x + y + z)^3 = (x^3 + y^3 + z^3) + 3 (x + y) (x + z) (y + z)$ it follows that $0 = 9 - 3xyz$ and $xyz = 3$.2012-04-28
  • 1
    So far I'm the only one who's up-voted this question, although one answer has six votes in its favor and another has two.2012-04-28
  • 2
    @MichaelHardy Why do you think it should be necessarily upvoted?2012-04-28
  • 0
    @PeterTamaroff : Because it's a question worth answering.2012-04-29
  • 0
    @MichaelHardy I see. The question has already quite the upvotes. Seems many agree with you!2012-04-29

2 Answers 2

14

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.

  • 0
    I didn't know about the Newton-girard, and glad to know that. But it will be a little hard to understand it for highschool students.2012-04-28
  • 4
    The Newton-Girard identities are derived by simply generalising the trick you used in your solution. Take the polynomial equation (e.g.) $t^4-at^3+bt^2-ct+d=0$ having roots $w,x,y,z$. You immediately identify $a,b,c,d$ as symmetric polynomials, and these are related to the sums of powers by substituting the individual roots into the equation and adding the four together. So they are easy to derive and no more difficult to understand than what you already know.2012-04-28
  • 0
    This gives you a way to check your answer since you know the values of all the symmetric polynomials (s1=0, s2=-2, s3=3) so x, y, and z should be the roots of t^3-2t-3 (which you can look up on Wolfram Alpha). You can try adding the roots, their cubes, and their 5ths to ensure you get 0, 9, and 30 respectively.2012-04-28
  • 0
    @MarkBennet Thanks, I got it. It was derived from the similar way!2012-04-29
8

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.

  • 0
    Yes, this is good and it's the same way I solved. I wrote this question expecting for some other easy methods, but maybe it's the easiest way.2012-04-28
  • 0
    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