If α and β are roots of the equation, x^2-4 √2 k x+2 e^4 ln k-1=0 for some k, and α^2+β^2=66, then α^3+β^3 is equal to:

$x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$x2−42​kx+2e4lnk−1=0 or, $x^2-4 \sqrt{2} k x+2 k^4-1=0$x2−42​kx+2k4−1=0 $\alpha+\beta=4 \sqrt{2} k$α+β=42​k and $\alpha \cdot \beta=2 k^4-1$α⋅β=2k4−1 Squaring both sides, we get $$\begin{aligned} &(\alpha+\beta)^2=(4 \sqrt{2} k)^2 \\ &\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=32 k^2 \\ &66+2 \alpha \beta=32 k^2 \\ &66+2\left(2 k^4-1\right)=32 k^2 \\ &66+4 k^4-2=32 k^2 \\ &\Rightarrow 4 k^4-32 k^2+64=0 \\ &\text { or, } k^4-8 k^2+16=0 \\ &\Rightarrow\left(k^2\right)^2-8 k^2+16=0 \\ &\Rightarrow\left(k^2-4\right)\left(k^2-4\right)=0 \\ &\Rightarrow k^2=4, k^2=4 \\ &\Rightarrow k=\pm 2 \end{aligned}$$​(α+β)2=(42​k)2⇒α2+β2+2αβ=32k266+2αβ=32k266+2(2k4−1)=32k266+4k4−2=32k2⇒4k4−32k2+64=0 or, k4−8k2+16=0⇒(k2)2−8k2+16=0⇒(k2−4)(k2−4)=0⇒k2=4,k2=4⇒k=±2​ Now, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)$α3+β3=(α+β)(α2+β2−αβ) $$\therefore \alpha^3+\beta^3=(4 \sqrt{2} k)\left[66-\left(2 k^4-1\right)\right]$$∴α3+β3=(42​k)[66−(2k4−1)] Putting $k=-2,(k=+2$k=−2,(k=+2 cannot be taken because it does not satisfy the above equation) $\therefore \alpha^3+\beta^3=(4 \sqrt{2}(-2))\left[66-2(-2)^4-1\right]$∴α3+β3=(42​(−2))[66−2(−2)4−1] $\alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)=$α3+β3=(−82​)(66−32+1)= $(-8 \sqrt{2})(35)$(−82​)(35) $\therefore \alpha^3+\beta^3=-280 \sqrt{2}$∴α3+β3=−2802​