If $a^{4} + b^{4} = a^{2} b^{2}$ , then $(a^{6} + b^{6})$ equal to

0

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1

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a^{2} + b^{2}

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

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Solution

The correct option is A $0$
$f o r m u l a u s e d$
$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$g i v e n$
$a^{6} + b^{6}$
$\Rightarrow {(a^{2})}^{3} + {(b^{2})}^{3}$
$\Rightarrow (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$
$\Rightarrow (a^{2} + b^{2}) (0)$
$\Rightarrow 0$

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