JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 63

$e^{4\ln k}=k^4$, so $\alpha+\beta=4\sqrt2k$ and $\alpha\beta=2k^4-1$.

$\alpha^2+\beta^2=32k^2-4k^4+2=66$ gives $(k^2-4)^2=0$, so $k=2$ ($k>0$ for $\ln k$). Then $\alpha+\beta=8\sqrt2$, $\alpha\beta=31$, and
$$\alpha^3+\beta^3=(8\sqrt2)^3-3(31)(8\sqrt2)=1024\sqrt2-744\sqrt2=280\sqrt2.$$