Factorize the below equation:
$1 - 2 a - 2 b - 3 {(a + b)}^{2}$

(1 - 3 b) (a + b)

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(1 - 3 a - 3 b) (1 + a + b)

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(a - 3 b) (1 - a + b)

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(1 - a - 3 b) (1 + a - b)

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Solution

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

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

\frac{a + b}{a^{2 / 3} - a^{1 / 3} b^{1 / 3} + b^{2 / 3}} - \frac{a - b}{a^{2 / 3} + a^{1 / 3} b^{1 / 3} b^{2 / 3}} - \frac{a^{2 / 3} - b^{2 / 3}}{a^{1 / 3} - b^{1 / 3}} \cdot

1-2a-2b - 3(a+b)^ 2 factorize

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

\frac{a - b}{a} + \frac{1}{2} (\frac{a - b}{a})^{2} + \frac{1}{3} (\frac{a - b}{a})^{3} + = \dots

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