If $\alpha$ and $\beta$ are roots of the equation, $x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ for some $k$, and $\alpha^2+\beta^2=66$, then $\alpha^3+\beta^3$ is equal to:

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Aicharya · Accepted Answer

$x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ or, $x^2-4 \sqrt{2} k x+2 k^4-1=0$ $\alpha+\beta=4 \sqrt{2} k$ and $\alpha \cdot \beta=2 k^4-1$ Squaring both sides, we get $ \begin{aligned} &(\alpha+\beta)^2=(4 \sqrt{2} k)^2 \ &\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=32 k^2 \ &66+2 \alpha \beta=32 k^2 \ &66+2\left(2 k^4-1ight)=32 k^2 \ &66+4 k^4-2=32 k^2 \ &\Rightarrow 4 k^4-32 k^2+64=0 \ &	ext { or, } k^4-8 k^2+16=0 \ &\Rightarrow\left(k^2ight)^2-8 k^2+16=0 \ &\Rightarrow\left(k^2-4ight)\left(k^2-4ight)=0 \ &\Rightarrow k^2=4, k^2=4 \ &\Rightarrow k=\pm 2 \end{aligned} $ Now, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \betaight)$ $ 	herefore \alpha^3+\beta^3=(4 \sqrt{2} k)\left[66-\left(2 k^4-1ight)ight] $ Putting $k=-2,(k=+2$ cannot be taken because it does not satisfy the above equation) $	herefore \alpha^3+\beta^3=(4 \sqrt{2}(-2))\left[66-2(-2)^4-1ight]$ $\alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)=$ $(-8 \sqrt{2})(35)$ $	herefore \alpha^3+\beta^3=-280 \sqrt{2}$

If $\alpha$ and $\beta$ are roots of the equation, $x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ for some $k$, and…

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