Factorise x2+y2+z2-xy-yz-zx

Question

As omega ina comex number. In a simple way for school grade student

Answer 1

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

x²+y²+z²-xy-yz-zx=0.

 

However, the graph is a line x=y=z as can be seen by using the Cauchy-Schwarz inequality:

 

(a₁² + ... + a_n²)(b₁² + ... + b_n²) ≥ (a₁b₁ + ... + a_nb_n)²

 

With equality happening if for all 1 ≤ k ≤ n, a_k = b_k.

 

In this case, take (a₁,a₂,a₃) = (x,y,z) and (b₁,b₂,b₃) = (y,z,x).

 

(x² + y² + z²)(y² + z² + x²) ≥ (xy + yz + zx)²

 

|x² + y² + z²| ≥ |xy + yz + zx|

 

with equality only along the line x=y=z.

 

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