Question

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

• A. $-32\sqrt{2}$
• B. $280\sqrt{2}$
• C. $-280\sqrt{2}$
• D. $248\sqrt{2}$

Answer 1

Answer: (B)

$280\sqrt{2}$

$\alpha +\beta =4\sqrt{2}k,\alpha \beta =2k^4-1$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

$66=16\times 2k^2-4k^4+2$

$16=8k^2-k^4$

$k^4-8k^2+16=0$

$\therefore k^2=4$

$k=\pm 2$

Since $k>0,k=+2$

So, equation becomes $x^2-8\sqrt{2}x+31=0$

$(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )={\alpha }^3+{\beta }^3$.

${\alpha }^3+{\beta }^3=(8\sqrt{2}{)}^3-3(8\sqrt{2}\left)\right(31)$

$=280\sqrt{2}$.