Factorise : $27 a^{3} - b^{3} - 1 - 9 a b$

(3 a + b - 1) (9 a^{2} + b^{2} + 1 - 3 a b + 3 a - b)

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(3 a - b - 1) (9 a^{2} - b^{2} + 1 + 3 a b + 3 a - b)

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(3 a + b + 1) (9 a^{2} + b^{2} + 1 + 3 a b + 3 a - b)

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(3 a - b - 1) (9 a^{2} + b^{2} + 1 + 3 a b + 3 a - b)

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Solution

The correct option is D $(3 a - b - 1) (9 a^{2} + b^{2} + 1 + 3 a b + 3 a - b)$
$27 a^{3} - b^{3} - 1 - 9 a b$
$= {(3 a)}^{3} + {(- b)}^{3} + {(- 1)}^{3} - 3 \times 3 a \times (- b) \times (- 1)$
$= [3 a + (- b) + (- 1)] [{(3 a}^{2} + {(- b)}^{2} + {(- 1)}^{2} - 3 a \times (- b) - (- b) \times (- 1) - 3 a \times (- 1)]$
$= (3 a - b - 1) (9 a^{2} + b^{2} + 1 + 3 a b - b + 3 a)$
$[∵ a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)]$

Suggest Corrections

Similar questions

27 a^{3} + b^{3} + c^{3} - 9 a b c = (3 a + b + c) [9 a^{2} + b^{2} + c^{2} - 3 a b - b c - 3 a c]

a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

(a - 2 b + 5 c) (a - b) - (a - b - c) (2 a + 3 c) + (6 a + b) (2 c - 3 a - 5 b)

19 c a - 37 a b - 19 a^{2}

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

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

Simplify :
$3 a - 2 b - a b - (a - b + a b) + 3 a b + b - a$

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