If $\alpha,\beta$ are the roots of the equation $x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$.
Then value of $\displaystyle \alpha^{98}(\alpha^{12}-1)+\beta^{98}(\beta^{12}-1)$ is
From $\displaystyle x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$,we have
$\displaystyle \alpha+\beta=-3^{\frac{1}{4}}$ and $\displaystyle \alpha\beta=\sqrt{3}$
Now we have to find value of $\displaystyle (\alpha^{110}+\beta^{110})-(\alpha^{98}+\beta^{98})$
I did not understand how can I write above expression into sum and product of roots
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Best Answer
Let $\omega$ and $\omega^2$ be the roots of $x^2 + x + 1 = 0$. Then the roots of the equation $x^2 + 3^{1/4}x + 3^{1/2} = 0$ are $\alpha = 3^{1/4}\omega$ and $\beta = 3^{1/4}\omega^2$.
Using the fact that $\omega^3 = 1$, we have, $\alpha^{12} - 1 = 3^3 \omega^{12} - 1 = 27 - 1 = 26$. Similarly, we can also conclude $\beta^{12} - 1 = 26$, $\alpha^{98} = 3^{24.5} \omega^2$ and $\beta^{98} = 3^{24.5} \omega$. Lastly, by using the relation $\omega + \omega^2 = -1$, we can conclude that \begin{align*} \alpha^{98}(\alpha^{12} - 1) + \beta^{98}(\beta^{12} - 1) = -26 \cdot 3^{24} \cdot \sqrt{3}. \end{align*}