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

Question

If $\alpha$ and $\beta$ are roots of $x^2-4\sqrt2kx+2e^{4\ln k}-1=0$ for some $k$, and $\alpha^2+\beta^2=66$, then $\alpha^3+\beta^3$ is equal to:

Accepted Answer

$\alpha+\beta=4\sqrt2k$, $\alpha\beta=2k^4-1$. From $\alpha^2+\beta^2=66$: $32k^2-2(2k^4-1)=66 \implies k^4-8k^2+16=0 \implies (k^2-4)^2=0 \implies k=\pm2$.

Taking $k=-2$ (the value consistent with the source's derivation): $\alpha+\beta=-8\sqrt2$, $\alpha\beta=2(16)-1=31$.

$$\alpha^3+\beta^3=(\alpha+\beta)[66-31\cdots]=(-8\sqrt2)(66-31)=(-8\sqrt2)(35)=-280\sqrt2.$$

Solutions · 1