Factorise the following: $x^{6} - 10 x^{3} - 27$

(x^{2} - x - 3) (x^{4} + x^{3} + 4 x^{2} - 3 x + 9)

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(x^{3} - x - 3) (x^{3} + x^{3} + 4 x^{2} + 3 x + 9)

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(x^{2} - x - 3) (x^{4} + x^{3} - 3 x + 9)

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(x^{3} - x - 3) (x^{3} + x^{3} + 4 x^{2} - 3 x - 9)

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Solution

The correct option is A $(x^{2} - x - 3) (x^{4} + x^{3} + 4 x^{2} - 3 x + 9)$
Given, $x^{6} - 10 x^{3} - 27$
$= x^{6} - x^{3} - 27 - 9 x^{3}$
$= (x^{2})^{3} + (- x)^{3} + (- 3)^{3} - 3 (x^{2}) (- x) (- 3)$
Using, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - a c - b c)$
$= (x^{2} - x - 3) ((x^{2})^{2} + (- x)^{2} + (- 3)^{2} - (x^{2}) (- x) - (x^{2}) (- 3) - (- x) (- 3))$
$= (x^{2} - x - 3) (x^{4} + x^{2} + 9 + x^{3} + 3 x^{2} - 3 x)$
$= (x^{2} - x - 3) (x^{4} + x^{3} + 4 x^{2} - 3 x + 9)$

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