Factorize:
$(a + b)^{3} - a - b$

(a + 2 b) (2 a + b + 1) (a - b + 1)

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(a + b) (a + b + 1) (a + b - 1)

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(a - b) (a + 2 b + 1) (a + b + 1)

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(a + 2 b) (2 a + 3 b + 1) (a + b + 1)

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Solution

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

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Similar questions

tan (\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}) + tan (\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b}) = \frac{2 b}{a}

tan [\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}] + tan [\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b}] = \frac{2 b}{a}

Factorize 98 $(a + b)^{2} - 2$

$t a n^{- 1} a + t a n^{- 1} b$ , where a>0,b>0,ab>1, is equal to

t a n (\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b})

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