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

$\because$ $x^3$ + $y^3$ + $z^3$- 3xyz = (x + y + z)($x^2$ + $y^2$+$z^2$- xy - yz -zx)

$\therefore$ RHS = $\frac{1}{2}$(x + y + z)(2 $x^2$ + 2 $y^2$ + 2 $z^2$- 2xy -2yz -2zx)

= $\frac{1}{2}$(x + y + z)($x^2$ + $x^2$ +$y^2$ + $y^2$ +$z^2$ + $z^2$ - 2xy -2yz -2zx)

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

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

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