Roman C. answered • 06/18/16

Tutor

4.9
(636)
Masters of Education Graduate with Mathematics Expertise

If this had factored, both factors would be linear, so the graph of the following equation would have been 1 or 2 planes

x

^{2}+y^{2}+z^{2}-xy-yz-zx=0.However, the graph is a line x=y=z as can be seen by using the Cauchy-Schwarz inequality:

(a

_{1}^{2}+ ... + a_{n}^{2})(b_{1}^{2}+ ... + b_{n}^{2}) ≥ (a_{1}b_{1}+ ... + a_{n}b_{n})^{2}With equality happening if for all 1 ≤ k ≤ n, a

_{k}= b_{k}.In this case, take (a

_{1},a_{2},a_{3}) = (x,y,z) and (b_{1},b_{2},b_{3}) = (y,z,x).(x

^{2}+ y^{2}+ z^{2})(y^{2}+ z^{2}+ x^{2}) ≥ (xy + yz + zx)^{2}|x

^{2}+ y^{2}+ z^{2}| ≥ |xy + yz + zx|with equality only along the line x=y=z.

The conclusion, is that it doesn't factor in the ordinary sense.