If `tan alpha, tan beta, tangamma` are the roots of the equation, `x^3- (a + 1) x^2 + (b - a) x-b= 0, (b-a !=1)` Where `alpha + beta +gamma` lies between `0 & pi` then `alpha + beta +gamma` is equal to

Options

A. `pi//4`
B. `pi//2`
C. `3pi//4`
D. None of these

Answer

To solve the problem, we need to find the value of \( \alpha + \beta + \gamma \) given that \( \tan \alpha, \tan \beta, \tan \gamma \) are the roots of the cubic equation:

\[ x^3 - (a + 1)x^2 + (b - a)x - b = 0 \]

### Step 1: Identify the relationships from the roots of the cubic equation

From Vieta's formulas, we know:

1. The sum of the roots \( \tan \alpha + \tan \beta + \tan \gamma = a + 1 \)
2. The sum of the products of the roots taken two at a time \( \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha = b - a \)
3. The product of the roots \( \tan \alpha \tan \beta \tan \gamma = b \)

### Step 2: Use the tangent addition formula

We can use the tangent addition formula for three angles:

\[ \tan(\alpha + \beta + \gamma) = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} \]

### Step 3: Substitute the values from Vieta's formulas

Substituting the values we found from Vieta's into the tangent addition formula:

- The numerator becomes:
\[ \tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma = (a + 1) - b \]

- The denominator becomes:
\[ 1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha) = 1 - (b - a) \]

### Step 4: Simplify the expression

Now, we can simplify the expression:

- The numerator simplifies to:
\[ (a + 1 - b) \]

- The denominator simplifies to:
\[ (1 - b + a) \]

Thus, we have:

\[ \tan(\alpha + \beta + \gamma) = \frac{a + 1 - b}{1 - b + a} \]

### Step 5: Set the equation equal to 1

Given that \( \alpha + \beta + \gamma \) lies between \( 0 \) and \( \pi \), we can set:

\[ \tan(\alpha + \beta + \gamma) = 1 \]

This implies:

\[ \frac{a + 1 - b}{1 - b + a} = 1 \]

### Step 6: Solve for \( a \) and \( b \)

Cross-multiplying gives us:

\[ a + 1 - b = 1 - b + a \]

This simplifies to:

\[ a + 1 - b = a + 1 - b \]

This equation holds true for all values of \( a \) and \( b \) as long as \( b - a \neq 1 \).

### Step 7: Conclude the value of \( \alpha + \beta + \gamma \)

Since \( \tan(\alpha + \beta + \gamma) = 1 \), we conclude:

\[ \alpha + \beta + \gamma = \frac{\pi}{4} \]

### Final Answer

Thus, the value of \( \alpha + \beta + \gamma \) is:

\[ \boxed{\frac{\pi}{4}} \]