We are asked to simplify two expressions: (1) Given that $x$ is real and $x \ne 0$, simplify the expression $(x - \frac{1}{x})^3 + (x + \frac{1}{x})^3$. (2) Simplify $\frac{log_{10} 125}{log_{10} 25}$.

1. Problem Description

We are asked to simplify two expressions:

(1) Given that $x$ is real and $x \ne 0$, simplify the expression $(x - \frac{1}{x})^3 + (x + \frac{1}{x})^3$.

(2) Simplify $\frac{log_{10} 125}{log_{10} 25}$.

2. Solution Steps

(1) Simplify the expression $(x - \frac{1}{x})^3 + (x + \frac{1}{x})^3$.

Recall the formula for $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ and $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

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

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

Adding these two expressions:

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

(2) Simplify $\frac{log_{10} 125}{log_{10} 25}$.

Recall that $log_a b = \frac{log_c b}{log_c a}$.

Using this, we can write $\frac{log_{10} 125}{log_{10} 25} = log_{25} 125$.

Since $125 = 5^3$ and $25 = 5^2$, we can write

$log_{25} 125 = log_{5^2} 5^3$.

Recall that $log_{a^b} x^y = \frac{y}{b} log_a x$.

So, $log_{5^2} 5^3 = \frac{3}{2} log_5 5 = \frac{3}{2}(1) = \frac{3}{2}$.

3. Final Answer

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

(2) $\frac{3}{2}$