If $a^{2} + a b + b^{2} = 19, a^{4} + a^{2} b^{2} + b^{4} = 133,$ then the value of ab is

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Solution

The correct option is A 6
Given $a^{2} + a b + b^{2} = 19$
$\Rightarrow a^{2} + b^{2} = 19 - a b$
And $a^{4} + a^{2} b^{2} + b^{4} = 133$
$\Rightarrow a^{4} + b^{4} + 2 a^{2} b^{2} - a^{2} b^{2} = 133$
$\Rightarrow (a^{2} + b^{2})^{2} - a^{2} b^{2} = 133$
Put value $\Rightarrow a^{2} + b^{2} = 19 - a b$
$(19 - a b)^{2} - a^{2} b^{2} = 133$
$\Rightarrow 361 - 38 a b + a^{2} b^{2} - a^{2} b^{2} = 133$
$\Rightarrow 38 a b = 361 - 133 \Rightarrow 38 a b = 228$
$\Rightarrow a b = 6$

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