Solving an Algebraic Equation

Date: 12/21/95 at 6:57:2
From: Anonymous
Subject: 7th grade math

If x = 2y = 3z =1, then what is x(y-z)squared + y(z-x)squared 
+ z(x-y)squared?

The answer I ended up was 7/18, But my teacher said it was 1/3.  
I don't get how it is done.

Date: 12/23/95 at 18:16:38
From: Doctor Elise
Subject: Re: 7th grade math

Hi!

I'm going to write "^2" instead of "squared" to save space.  Here we go:

>If x=2y=3z=1, then what is x(y-z)square + y(z-x)square + z(x-y)square??

x  = 1
2y = 1, so y = 1/2
3z = 1, so z = 1/3

So, x(y-z)^2 = 1(1/2 -1/3)^2 = (1/6)^2 = 1/36, and 

y(z-x)^2 = 1/2(1/3 - 1)^2 = 1/2(-2/3)^2 = 1/2(4/9) = 4/18 = 2/9

and

z(x-y)^2 = 1/3(1-1/2)^2 = 1/3(1/2)^2 = 1/3(1/4) = 1/12

Finally,  1/36 + 2/9 + 1/12 = 1/36 + 8/36 + 3/36 = 12/36 = 1/3

Ta da!

-Doctor Elise,  The Geometry Forum