Q.164. Algebra proof.

Question 164.
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.

Answer 164.
x^3 + y^3 + z^3 - 3xyz
= (x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)
= (1/2) (x + y + z) [(x - y)^2 + (y - z)^2 + (z - x)^2] ... ( 1 )

x + y + z
= a^2 - bc + b^2 - ca + c^2 - ab
= (1/2) [(a - b)^2 + (b -c)^2 + (c - a)^2] ... ( 2 )

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] ... ( 3 )

Plugging results ( 2 ) and ( 3 ) into ( 1),
x^3 + y^3 + z^3 - 3xyz
= (1/2) * (1/2) [(a - b)^2 + (b -c)^2 + (c - a)^2] * (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}]^2
which is a perfect square.

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