Factorise $a^{6} - 7 a^{3} - 8.$

(a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)

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(a + 2) (a^{2} + a + 4) (a + 1) (a^{2} + a + 1)

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(a - 2) (a^{2} + 4) (a + 1) (a^{3} - a - 1)

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(a - 2) (a^{2} + a + 2) (a + 1) (a^{2} - a - 1)

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Solution

The correct option is A $(a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)$
$a^{6} - 7 a^{3} - 8$
$= (a^{3})^{2} - 7 a^{3} - 8$
Let $a^{3} = x$
Now,
$x^{2} - 7 x - 8$
$= x^{2} - 8 x + x - 8$
$= x (x - 8) + 1 (x - 8)$
$= (x - 8) (x + 1)$
$= (a^{3} - 8) (a^{3} + 1)$
$= (a^{3} - 2^{3}) (a^{3} + 1)$
We know that,
{ $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ }
{ $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ }
$= (a - 2) (a^{2} + 2 a + 2^{2}) (a + 1) (a^{2} - a + 1)$
$= (a - 2) (a^{2} + 2 a + 4) (a + 1) (a^{2} - a + 1)$

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