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If $\mathrm{a}$ and $\mathrm{b}$ are the roots of the equation $\mathrm{y}^2+\mathrm{y}+1=0$, then the value of $a^4+b^4+a^{-1} b^{-1}$ is
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The correct answer is:
$0$
Given the equation $y^2+y+1=0$
$\begin{aligned}
& \text { Now, } a^4+b^4+a^{-1} b^{-1}=a^4+b^4+\frac{1}{a b} \\
& =\left(a^2+b^2\right)^2-2 a^2 b^2+\frac{1}{1} \\
& =\left((a+b)^2-2 a b\right)^2-2 \times 1+1 \\
& =\left((-1)^2-2\right)^2-1=1-1=0
\end{aligned}$
$\begin{aligned}
& \text { Now, } a^4+b^4+a^{-1} b^{-1}=a^4+b^4+\frac{1}{a b} \\
& =\left(a^2+b^2\right)^2-2 a^2 b^2+\frac{1}{1} \\
& =\left((a+b)^2-2 a b\right)^2-2 \times 1+1 \\
& =\left((-1)^2-2\right)^2-1=1-1=0
\end{aligned}$
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