The factorisation of $a^{4} + b^{4} - 7 a^{2} b^{2}$ is equal to .

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

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

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

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

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Solution

The correct option is C $(a^{2} + b^{2} + 3 a b) (a^{2} + b^{2} - 3 a b)$
The given expression is $a^{4} + b^{4} - 7 a^{2} b^{2}$
This can be written as
$a^{4} + b^{4} + 2 a^{2} b^{2} - 2 a^{2} b^{2} - 7 a^{2} b^{2}$
= $a^{4} + b^{4} + 2 a^{2} b^{2} - 9 a^{2} b^{2}$
= $(a^{2})^{2} + (b^{2})^{2} + 2 a^{2} b^{2} - 9 a^{2} b^{2}$
= $(a^{2} + b^{2})^{2} - 9 a^{2} b^{2}$
= $(a^{2} + b^{2})^{2} - (3 a b)^{2}$
= $(a^{2} + b^{2} - 3 a b) (a^{2} + b^{2} + 3 a b)$

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