More x(s), more difficult, but not impossible, although.

Since the $RHS = \dfrac {70737}{256} \gg 1$

$\Rightarrow x^4 \approx \dfrac {70737}{256} \quad \Rightarrow x \approx \sqrt [4] {\frac {70737}{256}} = 4.0771 \approx \boxed{4}$

I saw the denominator of the RHS as 256, and so I guessed four, but here's the solution.

Let $a=x+\frac {1}{x}$

$x^{2}+\frac {1}{x^{2}} = \left (x+\frac {1}{x} \right)^{2} - 2 = a^{2} - 2$

$x^{4} + \frac {1}{x^{4}} = \left(x^{2} + \frac {1}{x^{2}} \right)^{2} - 2 = \left ( \left ( x+\frac {1}{x} \right)^{2} - 2 \right)^{2} - 2 = (a^{2} - 2)^{2} - 2$

This you then substitute in the given equation, which becomes a degree four equation in $a$ , and once we have the value of $a$ , we can form a quadratic equation in $x$ and obtain the answer, $\boxed{4}$ .

To quickly deduct the answer I jst saw the lrgest denominatr in the LHS ...which is in fact the LCM

Clearly it is $x^{4}$ and is equal to 256

Now We see $256^{1/4} = 4$

So the answer is 4