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If \(2 \alpha=-1-i \sqrt{3}\) and \(2 \beta=-1+i \sqrt{3}\), then \(5 \alpha^4+5 \beta^4+7 \alpha^{-1} \beta^{-1}\) is equal to
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\(2 \alpha=-1-i \sqrt{3}, 2 \beta=-1+i \sqrt{3}\)
\(\left(5 \alpha^4+5 \beta^4+7 \alpha^{-1} \beta^{-1}\right)\)
\(\begin{aligned}
& =5\left(\alpha^4+\beta^4\right)+\frac{7}{\alpha \beta} \\
& =5\left[\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\right]+\frac{7}{\alpha \cdot \beta} \\
& =5\left[\left\{\frac{1}{4}(2 \alpha+2 \beta)^2-2 \alpha \beta\right\}^2-2 \alpha^2 \beta^2\right]+\frac{7}{\alpha \cdot \beta} \\
& =5\left[\left(\frac{1}{4} \times 4-2\right)^2-2\right]+7=5[1-2]+7=2
\end{aligned}\)
\(\left(5 \alpha^4+5 \beta^4+7 \alpha^{-1} \beta^{-1}\right)\)
\(\begin{aligned}
& =5\left(\alpha^4+\beta^4\right)+\frac{7}{\alpha \beta} \\
& =5\left[\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\right]+\frac{7}{\alpha \cdot \beta} \\
& =5\left[\left\{\frac{1}{4}(2 \alpha+2 \beta)^2-2 \alpha \beta\right\}^2-2 \alpha^2 \beta^2\right]+\frac{7}{\alpha \cdot \beta} \\
& =5\left[\left(\frac{1}{4} \times 4-2\right)^2-2\right]+7=5[1-2]+7=2
\end{aligned}\)
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