Xét mẫu thức:

`(x - y)^2 + (y - z)^2 + (z - x)^2`

`= (x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)`

`= 2(x^2 + y^2 + z^2) - 2(xy + yz + zx)`

Từ `x + y + z = 0`, suy ra `(x + y + z)^2 = 0`

`=> x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0`

`=> 2(xy + yz + zx) = -(x^2 + y^2 + z^2)` 

Suy ra: `2(x^2 + y^2 + z^2) - [-(x^2 + y^2 + z^2)]`

`= 2(x^2 + y^2 + z^2) + (x^2 + y^2 + z^2)`

`= 3(x^2 + y^2 + z^2)`

Vì `x, y, z ne 0` và `x + y + z = 0`

Nên `x^2 + y^2 + z^2 > 0`

`=> Q = (x^2 + y^2 + z^2) / [3(x^2 + y^2 + z^2)] = 1/3`

Vậy `Q = 1/3`