Factorise: $a^{4} + 2 a^{2} b^{2} + b^{4}$

(a + b)^{2}

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(a - b)^{2}

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

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

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Solution

The correct option is D $(a^{2} + b^{2})^{2}$
$a^{4} + 2 a^{2} b^{2} + b^{4}$
$= (a^{2})^{2} + 2 a^{2} b^{2} + (b^{2})^{2}$
$= (a^{2} + b^{2})^{2}$

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Similar questions

If X = 1 - a^{2} and Y = 1 - b^{2}, then what is X^{2} + Y^{2} ?

Question 92 (xviii)

Factorise the following using the identity $a^{2} - b^{2} = (a + b) (a - b) .$

$a^{4} - (a - b)^{4}$

a^{4} + a^{2} b^{2} + b^{4}

a^{2} + b^{2} + a b

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}}

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

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