Question

2x^3 + 2y^3 + 2z^3 - 6xyz

Answer 1

2x3 + 2y3 + 2z3– 6 xyz = (x + y + z) {(x – y)2 + (y – z)2 + (z – x)2} Let us solve the given expression using the identity: 2(13)3 + 2(14)3 + 2(15)3 – 6 × 13 × 14 × 15 = (13 + 14 + 15) {(13 – 14)2 + (14 – 15)2 + (15 – 13)2} = 42 (1 + 1 + 4) = 252 => 2 (x3 +y3 +z3 -3 xyz) = (x +y +z) {(x2 +y2 -2xy) +(y2 +z2 -2yz) +(z2 +x2 -2xz)} => 2 (x3 +y3 +z3 -3 xyz) = (x +y +z) (x2 +y2 -2xy +y2 +z2 -2yz +z2 +x2 -2xz) => 2 (x3 +y3 +z3 -3 xyz) = (x +y +z) (2x2 +2y2 +2z2 -2xy -2yz -2zx) => 2 (x3 +y3 +z3 -3 xyz) = 2 {(x +y +z) (x2 +y2 +z2 -xy -yz -zx)} => 2 (x3 +y3 +z3 -3 xyz) = 2 (x3 +y3 +z3 -3xyz) [ using identity, a3 +b3 +c3 -3abc= (a +b +c) (a2 +b2 +c2 -ab -bc -ac) ]