The given equation appears to be written incorrectly. I assume you meant:

$x^3 - x - 24 = 0$

Step 1: Check for Rational Roots

Using the Rational Root Theorem, the possible rational roots are factors of $24$, which are:

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

Let's test $x = 3$:

$(3)^3 - 3 - 24 = 27 - 3 - 24 = 0$

So, $x = 3$ is a root.

Step 2: Polynomial Division

Now, divide $x^3 - x - 24$ by $(x - 3)$ using synthetic division.

3 & 1 & 0 & -1 & -24 \\ & & 3 & 9 & 24 \\ \hline & 1 & 3 & 8 & 0 \end{array}$$ The quotient is: $$x^2 + 3x + 8$$ ### Step 3: Solve $$x^2 + 3x + 8 = 0$$ Using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = 1$$, $$b = 3$$, and $$c = 8$$: $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(8)}}{2(1)}$$ $$x = \frac{-3 \pm \sqrt{9 - 32}}{2}$$ $$x = \frac{-3 \pm \sqrt{-23}}{2}$$ $$x = \frac{-3 \pm i\sqrt{23}}{2}$$ ### Final Answer The three solutions are: $$x = 3, \quad x = \frac{-3 + i\sqrt{23}}{2}, \quad x = \frac{-3 - i\sqrt{23}}{2}$$ Would you like a detailed breakdown of any step? --- Here are five related questions for more practice: 1. Solve $$x^3 - 2x^2 - 5x + 6 = 0$$. 2. Solve $$x^3 - 4x^2 + 5x - 2 = 0$$ using the Rational Root Theorem. 3. Factorize $$x^3 + 6x^2 + 11x + 6$$. 4. Find all roots of $$x^3 - 7x + 6 = 0$$. 5. Solve $$x^3 - x^2 - 4x + 4 = 0$$ by synthetic division. ### Tip: When solving cubic equations, always check for rational roots first before using the quadratic formula or factoring!