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

The equation is x^2 - 4*sqrt(2)*k*x + 2*k^4 - 1 = 0. Sum of roots (a+b) = 4*sqrt(2)*k and product (ab) = 2*k^4 - 1. Given a^2+b^2 = (a+b)^2 - 2ab = 66, we have 32*k^2 - 2(2*k^4 - 1) = 66, which simplifies to -4*k^4 + 32*k^2 - 64 = 0, or k^4 - 8*k^2 + 16 = 0. This is (k^2 - 4)^2 = 0, so k^2 = 4. Then a+b = 8*sqrt(2) and ab = 2(16)-1 = 31. a^3+b^3 = (a+b)((a+b)^2 - 3ab) = 8*sqrt(2)(128 - 93) = 8*sqrt(2)(35) = 280*sqrt(2).