Factorise a-a3-b3-b is A- a-b1-a2-b2-a bB- a-b1-a2-b-a bC- a-b1-a2-b2-a bD- a-b1-a2-b2-a b

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Standard IX

Mathematics

Algebraic Identities

Factorise a-a...

Question

Factorise $a - a^{3} + b^{3} - b$ is _______

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

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

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

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

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Solution

The correct option is A $(a - b) (1 - a^{2} - b^{2} - a b)$
$a - a^{3} + b^{3} - b$
$= (a - b) + b^{3} - a^{3}$
$= (a - b) + [(b - a) (a^{2} + b^{2} + a b)]$
{ $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ }
$= (a - b) - (a - b) (a^{2} + b^{2} + a b)$
$= (a - b) (1 - a^{2} - b^{2} - a b)$

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Factorise a−a3+b3−b is _______

The correct option is A (a−b)(1−a2−b2−ab)a−a3+b3−b =(a−b)+b3−a3 =(a−b)+[(b−a)(a2+b2+ab)] { a3−b3=(a−b)(a2+ab+b2)} =(a−b)−(a−b)(a2+b2+ab) =(a−b)(1−a2−b2−ab)

Factorise $a - a^{3} + b^{3} - b$ is _______

The correct option is A $(a - b) (1 - a^{2} - b^{2} - a b)$
$a - a^{3} + b^{3} - b$
$= (a - b) + b^{3} - a^{3}$
$= (a - b) + [(b - a) (a^{2} + b^{2} + a b)]$
{ $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ }
$= (a - b) - (a - b) (a^{2} + b^{2} + a b)$
$= (a - b) (1 - a^{2} - b^{2} - a b)$