If $\alpha$ and $\beta$ are the roots of the equation $x^2-2 x+4=0$, then $\alpha^9+\beta^9$ is equal to

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Aicharya · Accepted Answer

Given quadratic equation is $ x^2-2 x+4=0 $ whose roots are $\alpha$ and $\beta$. $ 	herefore \quad \alpha+\beta=2 	ext { and } \alpha \beta=4 $ Now, $ \begin{aligned} & \alpha^9+\beta^9=\left(\alpha^3ight)^3+\left(\beta^3ight)^3 \ & =\left(\alpha^3+\beta^3ight)\left(\alpha^6+\beta^6-\alpha^3 \beta^3ight) \ = & (\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2ight)\left[\left(\alpha^2ight)^3+\left(\beta^2ight)^3-\alpha^3 \beta^3ight] \ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \betaight] \ & \quad\left[\left(\alpha^2+\beta^2ight)\left(\alpha^4+\beta^4-\alpha^2 \beta^2ight)-\alpha^3 \beta^3ight] \ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \betaight]\left[\left\{(\alpha+\beta)^2-2 \alpha \betaight\}ight. \ & \left.\left\{\left(\alpha^2+\beta^2ight)^2-3 \alpha^2 \beta^2ight\}-\alpha^3 \beta^3ight] \ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \betaight]\left[\left\{(\alpha+\beta)^2-2 \alpha \betaight\}ight. \ & \left.\quad\left[\left\{(\alpha+\beta)^2-(2 \alpha \beta)ight]^2-3 \alpha^2 \beta^2ight\}-\alpha^3 \beta^3ight] \ = & 2[4-12]\left[\{4-8\}\left\{(4-8)^2-48ight\}-64ight] \end{aligned} $ [from Eq. (i)] $ \begin{aligned} & =2(-8)\{(-4)(-32)(-64)\} \ & =2(-8)(128-64) \ & =2(-8)(64)=-2^{10} \end{aligned} $

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2 x+4=0$, then $\alpha^9+\beta^9$ is equal to

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