The factors of $8 a^{3} + b^{3} - 6 a b + 1$ are:

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

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

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

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

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Solution

The correct option is D $(2 a + b + 1) (4 a^{2} + b^{2} + 1 - 2 a b - b - 2 a)$
Given, $8 a^{3} + b^{3} - 6 a b + 1$ .

Then,
$8 a^{3} + b^{3} - 6 a b + 1$

$= 8 a^{3} + b^{3} - (3 \times 2 a \times b \times 1) + 1^{3}$

$= (2 a)^{3} + b^{3} + 1^{3} - (3 \times 2 a \times b \times 1)$ .

Since, $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ .

$= (2 a + b + 1) [(2 a)^{2} + b^{2} + 1^{2} - (2 a \times b) - (b \times 1) - (1 \times 2 a)] .$

$= (2 a + b + 1) (4 a^{2} + b^{2} + 1 - 2 a b - b - 2 a)$ .

Hence, the factors of $8 a^{3} + b^{3} - 6 a b + 1$ are $(2 a + b + 1) (4 a^{2} + b^{2} + 1 - 2 a b - b - 2 a)$ .

Therefore, option $C$ is correct.

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Similar questions

The factors of $8 a^{3} + b^{3} - 6 a b + 1$ are:

8 a^{3} - b^{3} - 4 a x + 2 b x

(2 a + b)^{2} - 6 a - 3 b - 4

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

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