Verify that \(x^3 + y^3 + z^3 - 3xyz = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)

We know that, \(x^3 + y^3 + z^3 - 3xyz \)
\(= (x + y + z)[x^2 + y^2 + z^2 - xy - yz - zx]\)
\( = \frac{1}{2} (x + y + z)[2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx]\)
\( = \frac{1}{2} (x + y + z)[x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx]\)
\( = \frac{1}{2} (x + y + z)[x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx]\)
\( = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)