If ax + cy + bz = X, cx + by + az = Y, bx + ay + cz = Z, show that :
$(a^{3} + b^{3} + c^{3} - 3 a b c) (x^{3} + y^{3} + z^{3} - 3 x y z) = (X^{3} + Y^{3} + Z^{3} - 3 X Y Z)$

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Solution

$(a^{3} + b^{3} + c^{3} - 3 a b c) (x^{3} + y^{3} + z^{3} - 3 x y z)$
$(a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) \times (x + y + z) (x^{2} + y^{2} + z^{2} - y z - z x - x y)$ ...(1)
Now $(a + b + c) (x + y + z) = [(a x + c y + b z) + (c x + b y + a z) + (b x + a y + c z)]$
= X + Y + Z ...(2)
And by part (1) we have
$(a^{2} + b^{2} + c^{2} - b c - c a - a b) (x^{2} + y^{2} + z^{2} - y z - z x - x y)$
$= X^{2} + Y^{2} + Z^{2} - Y Z - Z X - X Y$ ........(3)
Substituting from (2) and (3) in (1). we get
$(a^{3} + b^{3} + c^{3} - 3 a b c) (x^{3} + y^{3} + z^{3} - 3 x y z)$
$= (X + Y + Z) (X^{2} + Y^{2} + Z^{2} - Y Z - Z X - X Y)$
$= X^{3} + Y^{3} + Z^{3} - 3 X Y Z$

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