SOLUTION TO PROBLEM 6

As a small step toward simplifying things, let

$$\frac{1}{a-b}=x,\qquad \frac{1}{b-c}=y,\qquad \frac{1}{c-a}=z.$$

Then

$$\frac{1}{x}=a-b,\qquad \frac{1}{y}=b-c,\qquad \frac{1}{z}=c-a,$$

and it is pretty obvious that

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.$$

Clearing of fractions gives yz+zx+xy=0. Now, the expression we are concerned about is

x²+y²+z²,

which, when mentioned in the same breath with xy+yz+zx, can't help but bring to mind the basic expansion

(x+y+z)²=x²+y²+z²+2(xy+yz+zx).

Since xy+yz+zx=0, this gives

x²+y²+z²=(x+y+z)²,

a perfect square!

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Oct 5 08:35:05 CDT 2001