Part 9

Algebra Level 4

$\begin{aligned} m&=&\sqrt[3]{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}-1} \\ n&=&\sqrt[3]{\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}+2} \end{aligned}$

Let the values of $m$ and $n$ be described above. Compute $2(20m+6n)^{2}-32$ .

The answer is 2016.

1 solution

Rohit Udaiwal
Jan 12, 2016

Note that $\sqrt{3\pm2\sqrt2}=\sqrt{(\sqrt2\pm1)^2}=\sqrt2\pm1$ and similarly $\sqrt{17+12\sqrt2}=\sqrt{(3\pm2\sqrt2)^2}=3\pm2\sqrt2$ . $m=\sqrt[3]{\sqrt2+1-\sqrt2+1-1}=\sqrt[3]{1}=1 \\ n=\sqrt[3]{3+2\sqrt2+3-2\sqrt2+3}=\sqrt[3]{8}=2 \\ \implies 2(20m+6n)^{2}-32=2(20+12)^2-32=2016.$

Part 9

The answer is 2016.

1 solution

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