Verify that: x^3+y^3+z^3–3xyz = (1/2) (x+y+z)[(x–y)^2+(y–z)^2+(z–x)^2]

Solution:

We know that,

x3+y3+z3−3xyz = (x+y+z)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]

= (1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]