Factorise:
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3}$

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Solution

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z H e r e, x = (a - 3 b), y = (3 b - c), z = (c - a)$
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3} = (a - 3 b + 3 b - c + c - a) [{(a - 3 b)}^{2} + {(3 b - c)}^{2} + {(c - a)}^{2} - (a - 3 b) (3 b - c) - (3 b - c) (c - a) - (c - a) (a - 3 b)] + 3 (a - 3 b) (3 b - c) (c - a) = 0 + 3 (a - 3 b) (3 b - c) (c - a) = 3 (a - 3 b) (3 b - c) (c - a)$

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Similar questions

$(a - 3 b)^{3} + (3 b - c)^{3} + (c - a)^{3}$

Factorize $(a - 3 b)^{3} + (3 b - c)^{3} + (c - a)^{3}$

{(b + c)}^{3} + {(c + a)}^{3} + {(a + b)}^{3} - 3 (b + c) (c + a) (a + b) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)

Factorise:

$a - b - a^{3} + b^{3}$

a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3}

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