Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x+y+z=0$

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

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

$xy+yz+zx=?$

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?

share|improve this question
    
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. –  Ross Millikan Apr 28 '12 at 17:06
    
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 $xyz = 3$. –  TMM Apr 28 '12 at 17:13
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. –  Michael Hardy Apr 28 '12 at 17:48
2  
@MichaelHardy Why do you think it should be necessarily upvoted? –  Pedro Tamaroff Apr 28 '12 at 17:49
    
@PeterTamaroff : Because it's a question worth answering. –  Michael Hardy Apr 29 '12 at 0:08

2 Answers 2

up vote 14 down vote accepted

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.

share|improve this answer
    
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. –  Gobi Apr 28 '12 at 17:30
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. –  Mark Bennet Apr 28 '12 at 17:54
    
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. –  dspyz Apr 28 '12 at 20:40
    
@MarkBennet Thanks, I got it. It was derived from the similar way! –  Gobi Apr 29 '12 at 1:05

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.

share|improve this answer
    
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. –  Gobi Apr 28 '12 at 17:28
    
The advantage is that it keeps the manipulations simple and reduces the opportunity for error. So it is a practical method for simple cases. –  Mark Bennet Apr 28 '12 at 17:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.