If `alpha` is the only real root of the equation `x^3 + bx^2 + cx + 1 = 0 (b < c)`, then the value of `tan^-1 alpha+ tan^-1 (alpha^-1)` is equal to :

Explanation

To solve the problem, we need to find the value of \( \tan^{-1} \alpha + \tan^{-1} \left( \frac{1}{\alpha} \right) \) given that \( \alpha \) is the only real root of the equation \( x^3 + bx^2 + cx + 1 = 0 \) with the condition \( b < c \).

### Step-by-step Solution:

1. **Understanding the Function**:
We start with the function defined by the equation:
\[ f(x) = x^3 + bx^2 + cx + 1 \]
We need to analyze this function to find the nature of its roots.

2. **Behavior of the Function**:
- As \( x \to -\infty \), \( f(x) \to -\infty \).
- As \( x = 0 \), \( f(0) = 1 \).
- As \( x \to \infty \), \( f(x) \to \infty \).

This indicates that the function crosses the x-axis at least once.

3. **Finding the Root**:
Since it is given that \( \alpha \) is the only real root, and \( b < c \), we conclude that \( \alpha \) must be negative (as the function only crosses the x-axis once).

4. **Using the Identity**:
We know the identity:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \quad \text{if } xy < 1 \]
Here, we will use \( x = \alpha \) and \( y = \frac{1}{\alpha} \).

5. **Calculating the Sum**:
We have:
\[ \tan^{-1} \alpha + \tan^{-1} \left( \frac{1}{\alpha} \right) = \tan^{-1} \left( \frac{\alpha + \frac{1}{\alpha}}{1 - \alpha \cdot \frac{1}{\alpha}} \right) \]
Simplifying the denominator:
\[ 1 - \alpha \cdot \frac{1}{\alpha} = 1 - 1 = 0 \]
This indicates that the sum approaches \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \) depending on the sign of \( \alpha \).

6. **Considering the Sign of \( \alpha \)**:
Since \( \alpha < 0 \), we have:
\[ \tan^{-1} \alpha + \tan^{-1} \left( \frac{1}{\alpha} \right) = \frac{\pi}{2} - \pi = -\frac{\pi}{2} \]

### Final Result:
Thus, the value of \( \tan^{-1} \alpha + \tan^{-1} \left( \frac{1}{\alpha} \right) \) is:
\[ \boxed{-\frac{\pi}{2}} \]