Question

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

Answer 1

It is known that,

x3 + y3 + z3 - 3xyz

= (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

= $\frac{1}{2}$ (x + y + z) [2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx]

=$\frac{1}{2}$ (x + y + z) [(x2 + y2 - 2xy) + (y2 + z2 - 2yz) + (x2 + z2 -2zx)]

= $\frac{1}{2}$ (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]