If the roots of the equation `x^(3) -10 x + 11 = 0` are u, v, and w, then the value of `3 cosec^(2) (tan^(-1) u + tan^(-1) v + tan^(-1 w)` is ____

Explanation

To solve the equation \( x^3 - 10x + 11 = 0 \) and find the value of \( 3 \csc^2(\tan^{-1} u + \tan^{-1} v + \tan^{-1} w) \), where \( u, v, w \) are the roots of the equation, we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is \( x^3 - 10x + 11 = 0 \). We can identify the coefficients:
- \( a = 1 \)
- \( b = 0 \)
- \( c = -10 \)
- \( d = 11 \)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
1. \( u + v + w = -\frac{b}{a} = -\frac{0}{1} = 0 \)
2. \( uv + vw + wu = \frac{c}{a} = \frac{-10}{1} = -10 \)
3. \( uvw = -\frac{d}{a} = -\frac{11}{1} = -11 \)

### Step 3: Set up the angles
Let:
- \( \tan^{-1} u = \alpha \)
- \( \tan^{-1} v = \beta \)
- \( \tan^{-1} w = \gamma \)

Thus, we need to find \( \tan(\alpha + \beta + \gamma) \).

### Step 4: Use the formula for tangent of sum
The formula for the tangent of the sum of three angles is:
\[ \tan(\alpha + \beta + \gamma) = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} \]
Substituting \( \tan \alpha = u \), \( \tan \beta = v \), \( \tan \gamma = w \):
\[ \tan(\alpha + \beta + \gamma) = \frac{u + v + w - uvw}{1 - (uv + vw + wu)} \]

### Step 5: Substitute the values from Vieta's
Using the values obtained from Vieta's:
- \( u + v + w = 0 \)
- \( uv + vw + wu = -10 \)
- \( uvw = -11 \)

We substitute these into the formula:
\[ \tan(\alpha + \beta + \gamma) = \frac{0 - (-11)}{1 - (-10)} = \frac{11}{1 + 10} = \frac{11}{11} = 1 \]

### Step 6: Find \( \alpha + \beta + \gamma \)
Since \( \tan(\alpha + \beta + \gamma) = 1 \), we have:
\[ \alpha + \beta + \gamma = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 7: Calculate \( 3 \csc^2(\alpha + \beta + \gamma) \)
Now, we need to calculate \( 3 \csc^2\left(\frac{\pi}{4}\right) \):
\[ \csc\left(\frac{\pi}{4}\right) = \frac{1}{\sin\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]
Thus,
\[ \csc^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2 \]
Therefore,
\[ 3 \csc^2\left(\frac{\pi}{4}\right) = 3 \times 2 = 6 \]

### Final Answer
The value of \( 3 \csc^2(\tan^{-1} u + \tan^{-1} v + \tan^{-1} w) \) is \( \boxed{6} \).