To solve the inequality $x^3 - x \geq 0$, let's follow these steps:

Step 1: Factor the inequality

First, rewrite the inequality:

$x^3 - x \geq 0$

Factor the expression on the left-hand side:

$x(x^2 - 1) \geq 0$

Now, notice that $x^2 - 1$ is a difference of squares:

$x(x - 1)(x + 1) \geq 0$

Step 2: Find the critical points

The critical points occur when any of the factors are equal to zero. Set each factor to zero:

$x = 0, \quad x - 1 = 0 \implies x = 1, \quad x + 1 = 0 \implies x = -1$

So, the critical points are $x = -1$, $x = 0$, and $x = 1$.

Step 3: Analyze the sign of the expression

We now analyze the sign of the expression $x(x - 1)(x + 1)$ in the intervals determined by the critical points: $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, \infty)$.

1. For $x \in (-\infty, -1)$:
   Each factor is negative, so the product is negative.

   $(-)(-)(-) = -$ Therefore, $x(x - 1)(x + 1) < 0$.

2. For $x \in (-1, 0)$:
   $x + 1$ is positive, while $x$ and $x - 1$ are negative, so the product is positive.

   $(-)(-)(+) = +$ Therefore, $x(x - 1)(x + 1) > 0$.

3. For $x \in (0, 1)$:
   $x$ is positive, but $x - 1$ is negative and $x + 1$ is positive, so the product is negative.

   $(+)(-)(+) = -$ Therefore, $x(x - 1)(x + 1) < 0$.

4. For $x \in (1, \infty)$:
   All the factors are positive, so the product is positive.

   $(+)(+)(+) = +$ Therefore, $x(x - 1)(x + 1) > 0$.

Step 4: Include critical points

At $x = -1$, $x = 0$, and $x = 1$, the expression equals zero:

$x(x - 1)(x + 1) = 0$

Step 5: Write the solution

The inequality $x(x - 1)(x + 1) \geq 0$ holds when the expression is either positive or zero. From our sign analysis, this happens in the intervals:

$x \in [-1, 0] \cup [1, \infty)$

Thus, the solution to the inequality is:

$x \in [-1, 0] \cup [1, \infty)$

Would you like further details or have any questions?

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Here are five related questions to explore further:

1. How would the solution change if the inequality were strict ($x^3 - x > 0$)?
2. What is the solution to the inequality $x^2 - x \geq 0$?
3. Can you solve a similar inequality for $x^4 - x \geq 0$?
4. How would you solve $x^3 - 2x \geq 0$?
5. Can you solve the equation $x^3 - x = 0$ for its exact roots?

Tip: When factoring polynomials, always check for differences of squares and common factors!