1 to 2 and 1 to 3 is simple, but what about 3 to 2?

Since $x^3 + \frac{1}{x^3} = 10 \sqrt{2}$ hence $x^ 6 + \frac{ 1}{x^6} = ( 10 \sqrt{2} ) ^2 - 2 = 198$ .

Let $x^2 + \frac{1}{x^2} = a$ , then we have $a^3 - 3a = 198$ . This gives us $(a-6)(a^2 + 6a + 33 ) = 0$ . We have a real root $a = 6$ , and 2 complex roots.

First, we let $x^2+\dfrac{1}{x^2}=a$ , We see from the properties Calvin listed that $x+\dfrac{1}{x}=\sqrt{a+2}$ , so $x^3+\dfrac{1}{x^3}=\sqrt{a+2}\cdot (a-1)=\sqrt{a^3-3a+2}$ .

Since $x^3+\dfrac{1}{x^3}=10\sqrt{2}$ , we have $\sqrt{a^3-3a+2}=10\sqrt{2}=\sqrt{200}$ , so $a^3-3a+2=200$

Using, RRT, we find that $a=6$ is a solution. Using synthetic division, we find that the resulting quadratic, $a^2+6a+33$ , has no real solutions. Thus $a=\boxed{6}$ is the solution we are looking for.

I'll admit that I did this "by eye": $10 \sqrt{2} \ = \ 2 \sqrt{2} \cdot 5 \ = \ a^3 - 3a \ = \ a \cdot (a^2 - 3)$ . Noting that $\sqrt{8} = 2 \sqrt{2}$ and $8 - 5 \ = \ 3$ , it becomes clear that $a^2 \ = \ 8$ . So $\ a^2 - 2 \ = \ 6 \ = \ x^2 + \frac{1}{x^2}$ .

Suppose $x<0$ , then the equation $x^3 + \frac {1}{x^3} = 10 \sqrt2$ can't be fulfilled, thus $x>0$ which makes $a$ positive real.

We have $a^3 - 3a = 10 \sqrt2$ , and we want to evaluate $a^2 - 2$ , let $b = a^2$ , so $b$ is real too.

$\begin{aligned} b \sqrt b - 3 \sqrt b & = & 10 \sqrt 2 \\ \sqrt b (b - 3) & = & 10 \sqrt 2 \\ b (b - 3)^2 & = & 200 \\ \end{aligned}$

With $200 = 5^2 \times 8$ , then $b = 8 \Rightarrow a^2 - 2 = b - 2 = \boxed{6}$