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

Answer 1

Answer: (B)

$x^{2} -4\sqrt{2}kx+2k^{4} - 1 = 0$
$\alpha+\beta = 2\sqrt{2}k$
$\alpha\beta = 2k^{4} -1$
$\Rightarrow \alpha ^{2}+\beta ^{2} = 66$
$\left(\alpha +\beta \right)^{2} -2\,\alpha \beta = 66$
$32\,k^{2}-2\,\left(2k^{4}-1\right) = 66$
$2 \left(2k^{4}\right) - 32\, k^{2} + 64 = 0$
$4 \left(k^{2}-4\right)^{2} = 0 \Rightarrow k^{2} = 4 \Rightarrow k = 2$
$ \alpha ^{3}+\beta ^{3} = \left(\alpha +\beta \right)^{3}-3\,\alpha\beta\left(\alpha +\beta \right)$
$= \left(\alpha +\beta \right)\left(\alpha^{2} +\beta^{2}-\alpha\beta \right)$
$= \left(8\sqrt{2}\right)\,\left(66-31\right) = 280\,\sqrt{2}$