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

1. Problem Description

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

2. Solution Steps

We are asked to simplify the expression $(x-\frac{1}{x})^3 + (x+\frac{1}{x})^3$.

Let $a = x-\frac{1}{x}$ and $b = x+\frac{1}{x}$. Then the expression is $a^3+b^3$. We know that

$$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$

Alternatively, we can use the formula

$$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$$

Also, we can expand the terms directly:

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

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

Adding them together, we have

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

$$= 2x^3 + \frac{6}{x} = 2x^3 + \frac{6}{x} = 2(x^3 + \frac{3}{x}) = 2(\frac{x^4+3}{x})$$

Therefore,

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

3. Final Answer

$\frac{2x^4+6}{x}$