If α and β are roots of the equation, x^2-4 2 k x+2 e^ 4 k -1=0 for...

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$$\begin{aligned} &(\alpha+\beta)^2=(4 \sqrt{2} k)^2 \\ &\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=32 k^2 \\ &66+2 \alpha \beta=32 k^2 \\ &66+2 \left(2 k^4-1\right)=32 k^2 \\ &66+4 k^4-2=32 k^2 \\ &\Rightarrow 4 k^4-32 k^2+64=0 \\ &\text{ or, } k^4-8 k^2+16=0 \\ &\Rightarrow \left(k^2\right)^2-8 k^2+16=0 \\ &\Rightarrow \left(k^2-4\right)\left(k^2-4\right)=0 \\ &\Rightarrow k^2=4, k^2=4 \\ &\Rightarrow k=\pm 2 \end{aligned}$$Now,$\alpha^3+\beta^3=(\alpha+\beta) \left(\alpha^2+\beta^2-\alpha \beta\right) \therefore \alpha^3+\beta^3=(4 \sqrt{2} k) \left[66-\left(2 k^4-1\right)\right] Puttingk=-2,(k=+2$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] \alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)= (-8 \sqrt{2})(35) \therefore \alpha^3+\beta^3=-280 \sqrt{2}