Question
Factorise: $(a-3 b)^3+(3 b-c)^3+(c-a)^3$
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Answer
We know $x^3 + y^3 + z^3 - 3xyz$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) x^3 + y^3 + z^3$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$
Here, $x = (a - 3b), y = (3b - c), z$
$= (c - a) (a - 3b)^3 + (3b - c)^3 + (c - a)^3$
$= (a - 3b + 3b - c + c - a) [(a - 3b)^2 + (3b - c)^2 + (c - a)^2 - (a - 3b)(3b - c) $$- (3b - c)(c - a) - (c - a)(a - 3b)] + 3(a - 3b)(3b - c)(c - a)$
$= 0 + 3(a - 3b)(3b - c)(c - a)$
$= 3(a - 3b)(3b - c)(c - a)$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) x^3 + y^3 + z^3$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$
Here, $x = (a - 3b), y = (3b - c), z$
$= (c - a) (a - 3b)^3 + (3b - c)^3 + (c - a)^3$
$= (a - 3b + 3b - c + c - a) [(a - 3b)^2 + (3b - c)^2 + (c - a)^2 - (a - 3b)(3b - c) $$- (3b - c)(c - a) - (c - a)(a - 3b)] + 3(a - 3b)(3b - c)(c - a)$
$= 0 + 3(a - 3b)(3b - c)(c - a)$
$= 3(a - 3b)(3b - c)(c - a)$
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