If x + (1)/(x) = sqrt(3), find the value of (x8 + 1)/(x4).

Given the equation:

$$x + \frac{1}{x} = \sqrt{3}$$

Square both sides of the equation:

$$\left(x + \frac{1}{x}\right)^2 = (\sqrt{3})^2$$

Expand the left side using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$$x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = 3$$ $$x^2 + 2 + \frac{1}{x^2} = 3$$

Subtract 2 from both sides:

$$x^2 + \frac{1}{x^2} = 3 - 2$$ $$x^2 + \frac{1}{x^2} = 1$$

We have $x^2 + \frac{1}{x^2} = 1$.
Square both sides of this equation:

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

Expand the left side:

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

Subtract 2 from both sides:

$$x^4 + \frac{1}{x^4} = 1 - 2$$ $$x^4 + \frac{1}{x^4} = -1$$

The expression to evaluate is $\frac{x^8 + 1}{x^4}$.
We can split this fraction:

$$\frac{x^8 + 1}{x^4} = \frac{x^8}{x^4} + \frac{1}{x^4}$$

Simplify the first term:

$$\frac{x^8 + 1}{x^4} = x^{8-4} + \frac{1}{x^4}$$ $$\frac{x^8 + 1}{x^4} = x^4 + \frac{1}{x^4}$$

From Step 2, we found that $x^4 + \frac{1}{x^4} = -1$.
Therefore,

$$\frac{x^8 + 1}{x^4} = -1$$

The final answer is $-1$.