$x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ 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, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)$
$$
\therefore \alpha^3+\beta^3=(4 \sqrt{2} k)\left[66-\left(2 k^4-1\right)\right]
$$
Putting $k=-2,(k=+2$ cannot be taken because it does not satisfy the above equation)
$\therefore \alpha^3+\beta^3=(4 \sqrt{2}(-2))\left[66-2(-2)^4-1\right]$
$\alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)=$
$(-8 \sqrt{2})(35)$
$\therefore \alpha^3+\beta^3=-280 \sqrt{2}$