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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Aicharya · Accepted Answer

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

If \(2 \alpha=-1-i \sqrt{3}\) and \(2 \beta=-1+i \sqrt{3}\), then \(5 \alpha^4+5 \beta^4+7 \alpha^{-1}…

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