If $$x + y + z = 0$$, then what is the value of $$\frac{(3y^2 + x^2 + z^2)}{(2y^2 - xz)}?$$

Solution 1:
As the answer is independent of variables and so we can assume values for x,y and z an solve
let x=1,y=-1,z=0 therefore x+y+z=1-1+0=0
$$\frac{(3y^2 + x^2 + z^2)}{(2y^2 - xz)}$$
=$$\frac{(3(-1)^2 + 1^2 + 0^2)}{(2(-1)^2 - 1*(0))}$$
=$$\frac{4}{2}$$
=2
Solution 2:$$\frac{(3y^2 + x^2 + z^2)}{(2y^2 - xz)}$$=k
$$(3y^2 + x^2 + z^2)$$=$$k(2y^2 - xz)$$
$$x^2 + z^2+kxz$$=$$2ky^2-3y^2$$
We know x+y+z=0
we can see that for k=2
we get $$(x+z)^{2}=y^{2}$$
x+z+y=0
Therefore value of k=2