See a 3 + b 3 + c 3 - 3abc = (a+ b + c)(a 2 + b 2 + c 2 - ab - bc - ca). If we multiply by 2 and divide by 2 then on the R.H.S side (a + b + c) / 2 (2a 2 + 2b 2 + 2c 2 - 2ab - 2bc - 2ca) (a+b+c)/2 (a 2 + b 2 - 2ab + b 2 + c 2 - 2bc + c 2 + b 2 - 2bc) = (a+b+c)/2([a - b] 2 + [b - c] 2 + [c - a] 2) = 1/2 (a + b +c)([a - b] 2 + [b - c] 2 + [c - a] 2) hence proved.