The problem asks to find the domain of the function $y = \log(\frac{x^2 - 4}{x^3 - 8})$.

1. Problem Description

The problem asks to find the domain of the function $y = \log(\frac{x^2 - 4}{x^3 - 8})$.

2. Solution Steps

To find the domain of the function $y = \log(\frac{x^2 - 4}{x^3 - 8})$, we need to consider two conditions:

(1) The argument of the logarithm must be greater than

0. (2) The denominator must not be equal to

0.

First, let's factor the numerator and denominator:

$x^2 - 4 = (x - 2)(x + 2)$

$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

Now, we have:

$\frac{x^2 - 4}{x^3 - 8} = \frac{(x - 2)(x + 2)}{(x - 2)(x^2 + 2x + 4)}$

We can simplify this expression, but we need to consider the case where $x = 2$, which makes the denominator zero. Therefore, $x \ne 2$.

If $x \ne 2$, we have:

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

Now, we need to find when $\frac{x + 2}{x^2 + 2x + 4} > 0$.

The quadratic $x^2 + 2x + 4$ has no real roots because its discriminant is $2^2 - 4(1)(4) = 4 - 16 = -12 < 0$. Since the leading coefficient is positive, the quadratic is always positive, i.e., $x^2 + 2x + 4 > 0$ for all real $x$.

Thus, we only need to consider when $x + 2 > 0$, which means $x > -2$.

However, we must exclude the case where $x = 2$, as it makes the original denominator zero. So, the domain is $x > -2$ and $x \ne 2$.

Therefore, the domain is $(-2, 2) \cup (2, \infty)$.

3. Final Answer

$(-2, 2) \cup (2, \infty)$