Algebra

First off, rearrange the given equation. This is simply $x=\frac{-1}{x}$ . Multiplying both sides by $x$ , $x^2=-1$ , and thus $x=\pm i$ . Now let us evaluate $x^{16}$ . Remember, $i^n=i^{n \pmod{4}}$ . Thus, $i^{16}=i^0=1$ . Now, plugging this newfound value of $x^{16}$ into the desired equation, we find $1+1=\boxed{2}$ . Great problem @Victor Loh !

$x + \dfrac{1}{x} = 0$

$\dfrac{x^2 +1}{x} = 0$

$x^2 + 1 = 0$ (Since $x \not = 0$ )

$x^2 = -1$

$x = \pm i$

Putting the value of x in $x^{16} + \dfrac{1}{x^{16}}$ , we get,

$(\pm i)^{16} + \dfrac{1}{(\pm i)^{16}}$

$= (-1)^8 + \dfrac{1}{(-1)^8}$

$= 1 + \dfrac{1}{1}$

$= \boxed{2}$

We know that zero can't be a denominator, so let's solve the equation.

First, $x + \frac { 1 } { x } = 0 \rightarrow x^{2} + 1 = 0 \rightarrow x^{2} = -1 \Rightarrow x = \pm i$ .

So, the roots for the equation are $\pm i$ .

Then, let's find the value of $x^{16} + \frac { 1 } { x ^ { 16 } }$ .

Substituting $x = \pm i$ to the equation, we get $( \pm i ) ^ { 16 } + \frac { 1 } { ( \pm i ) ^ { 16 } }$ .

Then, we get $1 + 1$ .

Thus, the value is $\boxed { 2 }$ .

$\therefore \boxed { 2 }$ is the answer for the problem.