Let’s break down **how to find the roots of a cubic equation** in detail: ## 1. **General Form of a Cubic Equation** A general cubic equation looks like this: \[ ax^3 + bx^2 + cx + d = 0 \] where \( a \neq 0 \). --- ## 2. **Step 1: Reduce to a Depressed Cubic** First, divide both sides by \( a \) (if \( a \neq 1 \)): \[ x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = 0 \] Now, **eliminate the \( x^2 \) term** by substituting: \[ x = y - \frac{b}{3a} \] This gives: \[ y^3 + py + q = 0 \] where: \[ p = \frac{3ac - b^2}{3a^2}, \quad q = \frac{2b^3 - 9abc + 27a^2d}{27a^3} \] --- ## 3. **Step 2: Cardano’s Formula** For the depressed cubic \( y^3 + py + q = 0 \): - Compute the **discriminant**: \[ \Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 \] - Roots depend on the sign of \( \Delta \): --- ### **Case 1: \( \Delta > 0 \) (One real root, two complex conjugates)** \[ y = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} \] --- ### **Case 2: \( \Delta = 0 \) (All roots real, at least two equal)** - If \( p = 0 \) and \( q = 0 \): triple root at \( y = 0 \). - Otherwise, \[ y_1 = 2\sqrt[3]{-\frac{q}{2}} \] \[ y_2 = -\sqrt[3]{-\frac{q}{2}} \] (with multiplicities) --- ### **Case 3: \( \Delta < 0 \) (All roots real and unequal)** Let \( \theta = \arccos \left( -\frac{q}{2} \cdot \left( -\frac{3}{p} \right)^{3/2} \right) \). Then, \[ y_k = 2\sqrt{-\frac{p}{3}} \cos \left( \frac{\theta + 2\pi k}{3} \right), \quad k = 0, 1, 2 \] --- ## 4. **Step 3: Substitute Back to Find \( x \)** Once you have \( y \) values, recall: \[ x = y - \frac{b}{3a} \] --- ## 5. **Example** Let’s solve: \[ x^3 - 3x^2 + 3x - 1 = 0 \] - Here, \( a=1, b=-3, c=3, d=-1 \). First, substitute \( x = y + \frac{3}{3} = y + 1 \): \[ (x)^3 - 3(x)^2 + 3(x) - 1 = 0 \] \[ (y+1)^3 - 3(y+1)^2 + 3(y+1) - 1 = 0 \] Expand: \[ y^3 + 3y^2 + 3y + 1 - 3(y^2 + 2y + 1) + 3y + 3 - 1 = 0 \] \[ y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 + 3y + 3 - 1 = 0 \] \[ y^3 + 0y^2 + 0y + 0 = 0 \] \[ y^3 = 0 \implies y = 0 \] So \( x = 0 + 1 = 1 \) is a **triple root**. --- ## 6. **Summary Table** | Step | Action | |------|------------------------------------------------------| | 1 | Divide by \( a \), write in standard form | | 2 | Substitute \( x = y - \frac{b}{3a} \) | | 3 | Get depressed cubic: \( y^3 + py + q = 0 \) | | 4 | Compute discriminant \( \Delta \) | | 5 | Use Cardano’s formulas for roots | | 6 | Substitute back to get \( x \) | --- **If you have a specific cubic equation in mind, feel free to provide it, and I can walk you through the process step by step!**