We need to find the domain of the function $g(x) = \frac{3x-1}{x^3-1}$.

1. Problem Description

We need to find the domain of the function $g(x) = \frac{3x-1}{x^3-1}$.

2. Solution Steps

The domain of a rational function is all real numbers except for the values of $x$ that make the denominator equal to zero. Therefore, we need to find the values of $x$ for which $x^3-1 = 0$.

$x^3 - 1 = 0$

$x^3 = 1$

$x = \sqrt[3]{1}$

$x = 1$

Therefore, the denominator is zero when $x=1$. So, $x$ cannot be equal to

1.

Alternatively, we can factor the denominator:

$x^3 - 1 = (x-1)(x^2+x+1)$.

The first factor gives $x-1=0$, so $x=1$.

The second factor is $x^2+x+1$. We find the roots of this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=1$, and $c=1$.

$x = \frac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}$

$x = \frac{-1 \pm \sqrt{1-4}}{2}$

$x = \frac{-1 \pm \sqrt{-3}}{2}$

$x = \frac{-1 \pm i\sqrt{3}}{2}$

Since the roots of $x^2+x+1$ are complex numbers, only $x=1$ makes the denominator zero. Thus, $x$ cannot be equal to

1.

In interval notation, the domain is $(-\infty, 1) \cup (1, \infty)$.

3. Final Answer

The domain of $g(x)$ is $(-\infty, 1) \cup (1, \infty)$.