The problem is to simplify the expression $(\frac{1}{x} - x)^3 + (x + \frac{1}{x})^3$, given that $x$ is a real number and $x \neq 0$. We are also asked to simplify $\frac{\log_{10} 125}{\log_{10} 25}$.

1. Problem Description

The problem is to simplify the expression $(\frac{1}{x} - x)^3 + (x + \frac{1}{x})^3$, given that $x$ is a real number and $x \neq 0$. We are also asked to simplify $\frac{\log_{10} 125}{\log_{10} 25}$.

2. Solution Steps

First, let's simplify the expression $(\frac{1}{x} - x)^3 + (x + \frac{1}{x})^3$. We can use the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ or expand each term individually and add them together. Let's use the latter method.

Recall that $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ and $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Then,

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

$(x + \frac{1}{x})^3 = x^3 + 3x^2(\frac{1}{x}) + 3x(\frac{1}{x})^2 + (\frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}$.

Adding these two expansions, we have

$(\frac{1}{x} - x)^3 + (x + \frac{1}{x})^3 = (\frac{1}{x^3} - \frac{3}{x} + 3x - x^3) + (x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}) = 2(\frac{1}{x^3} + 3x) = \frac{2}{x^3} + 6x$.

However, this is not one of the given options. Let's re-evaluate with a different approach.

Let $a = \frac{1}{x} - x$ and $b = x + \frac{1}{x}$. Then $a+b = \frac{1}{x} - x + x + \frac{1}{x} = \frac{2}{x}$.

We want to compute $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 - 3ab)$.

Here, $ab = (\frac{1}{x} - x)(x + \frac{1}{x}) = 1 + \frac{1}{x^2} - x^2 - 1 = \frac{1}{x^2} - x^2$.

So $a^3+b^3 = \frac{2}{x}((\frac{2}{x})^2 - 3(\frac{1}{x^2} - x^2)) = \frac{2}{x}(\frac{4}{x^2} - \frac{3}{x^2} + 3x^2) = \frac{2}{x}(\frac{1}{x^2} + 3x^2) = \frac{2}{x^3} + 6x = 2(\frac{1}{x^3} + 3x)$.

The expression can be written as $2(3x + \frac{1}{x^3})$.

Next, let's simplify $\frac{\log_{10} 125}{\log_{10} 25}$.

We know that $125 = 5^3$ and $25 = 5^2$.

Then, using the change of base formula, $\frac{\log_{10} 125}{\log_{10} 25} = \frac{\log_{5} 125}{\log_{5} 25} = \frac{\log_{5} 5^3}{\log_{5} 5^2} = \frac{3\log_{5} 5}{2\log_{5} 5} = \frac{3}{2}$.

Alternatively, $\frac{\log_{10} 125}{\log_{10} 25} = \frac{\log_{10} 5^3}{\log_{10} 5^2} = \frac{3 \log_{10} 5}{2 \log_{10} 5} = \frac{3}{2}$.

3. Final Answer

For the first problem, the simplified expression is $2(3x + \frac{1}{x^3})$. However, none of the options match. Option (d) is $x^3 + \frac{3}{x}$. Let's re-evaluate $(x + \frac{1}{x})^3 + (\frac{1}{x} - x)^3$.

If we have to pick one of the provided options, let's see what is closest. We found $6x + \frac{2}{x^3}$, which is $2(3x + \frac{1}{x^3})$. It is possible that there's an error somewhere, and we must choose the closest expression, which would likely be $2(x^3 + \frac{1}{x^3})$ if it was offered. But if instead one of the initial expression should be $(x - \frac{1}{x})^3$, then the answer would be $2x^3 + \frac{6}{x}$. But given the expressions, I am uncertain.

For the second problem, the simplified expression is $\frac{3}{2}$.