Find the range of `lambda` for which equation `x^3+x^2-x-1-lambda=0` has `3` real solution.

Answer

To find the range of \(\lambda\) for which the equation \(x^3 + x^2 - x - 1 - \lambda = 0\) has 3 real solutions, we can follow these steps:

### Step 1: Define the function
Let \(f(x) = x^3 + x^2 - x - 1 - \lambda\). We want to find the values of \(\lambda\) such that \(f(x) = 0\) has 3 real solutions.

### Step 2: Find the derivative
To determine the critical points, we need to find the derivative of \(f(x)\):
\[ f'(x) = \frac{d}{dx}(x^3 + x^2 - x - 1 - \lambda) = 3x^2 + 2x - 1 \]

### Step 3: Set the derivative to zero
Now, we set the derivative equal to zero to find the critical points:
\[ 3x^2 + 2x - 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \]
This gives us:
\[ x_1 = \frac{2}{6} = \frac{1}{3}, \quad x_2 = \frac{-6}{6} = -1 \]

### Step 4: Analyze the critical points
We have two critical points \(x = -1\) and \(x = \frac{1}{3}\). To determine the nature of these critical points, we can evaluate the second derivative:
\[ f''(x) = 6x + 2 \]
Evaluating at the critical points:
- For \(x = -1\):
\[ f''(-1) = 6(-1) + 2 = -6 + 2 = -4 \quad (\text{local maximum}) \]
- For \(x = \frac{1}{3}\):
\[ f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) + 2 = 2 + 2 = 4 \quad (\text{local minimum}) \]

### Step 5: Find the function values at critical points
Now, we need to find the values of the function at these critical points:
- For \(x = -1\):
\[ f(-1) = (-1)^3 + (-1)^2 - (-1) - 1 - \lambda = -1 + 1 + 1 - 1 - \lambda = 0 - \lambda = -\lambda \]
- For \(x = \frac{1}{3}\):
\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right) - 1 - \lambda \]
Calculating each term:
\[ = \frac{1}{27} + \frac{1}{9} - \frac{1}{3} - 1 - \lambda = \frac{1}{27} + \frac{3}{27} - \frac{9}{27} - \frac{27}{27} - \lambda = \frac{1 + 3 - 9 - 27}{27} - \lambda = \frac{-32}{27} - \lambda \]

### Step 6: Set conditions for three real solutions
For \(f(x) = 0\) to have three real solutions, we need:
1. \(f(-1) > 0\) (local maximum must be above the x-axis):
\[ -\lambda > 0 \implies \lambda < 0 \]
2. \(f\left(\frac{1}{3}\right) < 0\) (local minimum must be below the x-axis):
\[ \frac{-32}{27} - \lambda < 0 \implies \lambda > -\frac{32}{27} \]

### Step 7: Combine the conditions
Combining these inequalities, we find:
\[ -\frac{32}{27} < \lambda < 0 \]

### Final Result
The range of \(\lambda\) for which the equation \(x^3 + x^2 - x - 1 - \lambda = 0\) has 3 real solutions is:
\[ \boxed{\left(-\frac{32}{27}, 0\right)} \]