Q2014/092) If a, b, c ϵ R, $x = a² - bc$, $y = b² - ca$, $z = c² - ab$, then prove that, $x³ + y³ + z³ - 3xyz$ is a perfect square

we have

$x^3 + y^3 + z^3 - 3xyz$
= $(x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)$
= $\dfrac{1}{2} (x + y + z) [(x - y)^2 + (y - z)^2 + (z - x)^2] \cdots(1)$... ( 1 )

$x + y + z$
= $a^2 - bc + b^2 - ca + c^2 - ab$
= $\dfrac{1}{2}[(a - b)^2 + (b -c)^2 + (c - a)^2] \cdots (2)$

further

$x - y$
= $(a^2 - bc) - (b^2 - ca)$
= $a^2 - b^2 + ca - bc$
= $(a - b) (a + b) + c (a - b)$
= (a + b + c) (a - b)

Similarly,
$y - z = (a + b + c) (b - c)$ and
$z - x = (a + b + c) (c - a)$

=> $(x - y)^2 + (y - z)^2 + (z - x)^2$
= $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)\cdots ( 3 )$

Plugging results ( 2 ) and ( 3 ) into ( 1),

$x^3 + y^3 + z^3 - 3xyz$
= $\dfrac{1}{2} * \dfrac{1}{2}((a - b)^2 + (b -c)^2 + (c - a)^2) *$

            $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)$
= $(\dfrac{1}{2} (a + b + c) ((a - b)^2 + (b - c)^2 + (c - a)^2))^2$

which is a perfect square.