If a^{4}+a^{2} b^{2}+b^{4}=8 and a^{2}+a b+b^ | vedclass

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(D) We know the algebraic identity: $a^{4}+a^{2} b^{2}+b^{4} = (a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$.Given $a^{4}+a^{2} b^{2}+b^{4} = 8$ and $a^{2}+a b+

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$1$

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(D) We know the algebraic identity: $a^{4}+a^{2} b^{2}+b^{4} = (a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$.Given $a^{4}+a^{2} b^{2}+b^{4} = 8$ and $a^{2}+a b+b^{2} = 4$.Substituting these values into the identity:$8 = 4(a^{2}-a b+b^{2})$$a^{2}-a b+b^{2} = 2$ ........$(1)$We are also given $a^{2}+a b+b^{2} = 4$ ........$(2)$Subtracting equation $(1)$ from equation $(2)$:$(a^{2}+a b+b^{2}) - (a^{2}-a b+b^{2}) = 4 - 2$$a^{2}+a b+b^{2}-a^{2}+a b-b^{2} = 2$$2ab = 2$$ab = 1$.

If $a^{4}+a^{2} b^{2}+b^{4}=8$ and $a^{2}+a b+b^{2}=4,$ then the value of $a b$ is

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