`x^2 + x + 1` is a factor of `ax^(3) + bx^2 + cx + d = 0`, then the real root of above equation is (`a,b,c,d in R`)

Answer

To find the real root of the equation \( ax^3 + bx^2 + cx + d = 0 \) given that \( x^2 + x + 1 \) is a factor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the roots of the quadratic factor**:
The quadratic equation \( x^2 + x + 1 = 0 \) can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \).
Plugging in these values, we get:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2} \]

2. **Determine the remaining real root**:
Since \( x^2 + x + 1 \) is a factor of the cubic polynomial, the remaining root (let's call it \( \alpha \)) must be real. The product of the roots of the cubic equation can be expressed as:
\[ \text{Product of roots} = \alpha \cdot \omega \cdot \omega^2 \]
We know that \( \omega \cdot \omega^2 = 1 \) (since \( \omega^3 = 1 \)). Therefore:
\[ \alpha \cdot 1 = \alpha \]

3. **Use Vieta's formulas**:
According to Vieta's formulas, the product of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ \text{Product of roots} = -\frac{d}{a} \]
Thus, we have:
\[ \alpha = -\frac{d}{a} \]

4. **Conclusion**:
Therefore, the real root of the equation \( ax^3 + bx^2 + cx + d = 0 \) is:
\[ \alpha = -\frac{d}{a} \]