We are tasked with solving the inequality:

$x^3 - 5x^2 - 6x < 0$

Step 1: Factor the expression

First, factor the left-hand side:

$x(x^2 - 5x - 6) < 0$

Next, factor the quadratic expression $x^2 - 5x - 6$:

$x(x - 6)(x + 1) < 0$

Step 2: Determine critical points

The critical points occur where the expression equals zero, which are:

$x = 0, \quad x = 6, \quad x = -1$

These divide the real number line into four intervals: $(-\infty, -1), \quad (-1, 0), \quad (0, 6), \quad (6, \infty)$

Step 3: Test the sign in each interval

We now determine the sign of $x(x - 6)(x + 1)$ in each of these intervals by picking a test point from each.

1. For $x \in (-\infty, -1)$, take $x = -2$: $(-2)((-2) - 6)((-2) + 1) = (-2)(-8)(-1) = -16 < 0$
2. For $x \in (-1, 0)$, take $x = -\frac{1}{2}$: $\left(-\frac{1}{2}\right)\left(-\frac{1}{2} - 6\right)\left(-\frac{1}{2} + 1\right) = \left(-\frac{1}{2}\right)(-6.5)(0.5) > 0$
3. For $x \in (0, 6)$, take $x = 1$: $(1)(1 - 6)(1 + 1) = 1(-5)(2) = -10 < 0$
4. For $x \in (6, \infty)$, take $x = 7$: $(7)(7 - 6)(7 + 1) = (7)(1)(8) = 56 > 0$

Step 4: Solution

The inequality is satisfied when the expression is negative, which occurs in the intervals:

$(-\infty, -1) \cup (0, 6)$

Since the inequality is strict (i.e., $< 0$ and not $\leq 0$), the critical points $x = -1$, $x = 0$, and $x = 6$ are not included.

Thus, the solution is:

$x \in (-\infty, -1) \cup (0, 6)$

Would you like more details or have any questions?

Here are some related questions:

1. How would the solution change if the inequality were $\leq 0$?
2. How do you factor quadratic expressions systematically?
3. What is the graphical interpretation of this inequality?
4. How do you solve inequalities involving higher powers of $x$?
5. What if the inequality were $x^3 - 5x^2 - 6x > 0$?

Tip: When solving inequalities, always check the sign in each interval created by the critical points.