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

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Solution

$a^{4} - 2 a^{2} b^{2} + b^{4} = a^{4} - a^{2} b^{2} - a^{2} b^{2} + b^{4} = a^{2} (a^{2} - b^{2}) - b^{2} (a^{2} - b^{2}) = (a^{2} - b^{2})^{2}$
As we know that
$x^{2} - y^{2} = (x + y) (x - y)$
$⟹ a^{4} - 2 a^{2} b^{2} + b^{4} = (a^{2} - b^{2})^{2} = (a - b)^{2} (a + b)^{2}$

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