Question

Prove that : 2x^3+2y^3+2z^3-6xyz=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] hence evaluate 2 (7)^3+2 (9)^3+2(13)^3-6 (7)(9)(13)

Answer 1

2x3 + 2y3 + 2z3 – 6xyz = 2 (x3 + y3 + z3 – 3xyz)

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

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

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

= (x + y + z) ((x – y)2 + (y – z)2 + (z – x)2)

so

2(7)3 + 2(9)3 + 2(13)3 – 6(7)(9)(13)

= (7 + 9 + 13) ((7 – 9)2 + (9 – 13)2 + (13 – 7)2)

= 29 ((–2)2 + (–4)2 + (6)2)

= 29 (4 + 16 + 36) = 29 × 56

= 1624