Factorise :
$8 x^{6} + 95 x^{3} + 1$

(x^{2} + 5 x + 1) (4 x^{4} - 10 x^{3} + 23 x^{3} - 5 x + 1)

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(2 x^{2} + 5 x + 1) (4 x^{4} - 10 x^{3} + 23 x^{3} - 5 x + 1)

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(2 x^{2} + x + 1) (4 x^{4} - 10 x^{3} + 23 x^{3} - 5 x + 1)

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(2 x^{2} + 5 x + 1) (4 x^{4} - 10 x^{2} + 23 x^{3} - 5 x + 1)

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Solution

$8 x^{6} + 95 x^{3} + 1$
Using the identity,
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
Now,
$8 x^{6} + 95 x^{3} + 1$
$= 8 x^{6} + 125 x^{3} + 1 - 30 x^{3}$
$= (2 x^{2})^{3} + (5 x)^{3} + (1)^{3} - 3 (2 x^{2}) (5 x) (1)$
Here,
$a = 2 x^{2}$
$b = 5 x$
$c = 1$ ,
Therefore,
$∴ (2 x^{2})^{3} + (5 x)^{3} + (1)^{3} - 3 (2 x^{2}) (5 x) (1)$
$= (2 x^{2} + 5 x + 1) ((2 x)^{2} + (5 x)^{2} + 1^{2} - 2 x^{2} (5 x) - 5 x (1) - ((2 x^{2}) (1)$
$= (2 x^{2} + 5 x + 1) (4 x^{4} + 25 x^{2} + 1 - 10 x^{3} - 5 x - 2 x^{2}$
$= (2 x^{2} + 5 x + 1) (4 x^{4} - 10 x^{3} + 23 x^{3} - 5 x + 1)$
Hence, Option B is correct.

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