Let a,b and c be the roots of `x ^(3) -x+1 =0,` then the ralue of `((1)/(a+1)+(1)/(b+1)+(1)/(c+1))` equals to :

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To find the value of \(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\) where \(a\), \(b\), and \(c\) are the roots of the equation \(x^3 - x + 1 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the cubic equation
The given cubic equation is:
\[ x^3 - x + 1 = 0 \]
From this, we can identify:
- \(a = 1\) (coefficient of \(x^3\))
- \(b = 0\) (coefficient of \(x^2\))
- \(c = -1\) (coefficient of \(x\))
- \(d = 1\) (constant term)

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
- The sum of the roots \(a + b + c = -\frac{b}{a} = -\frac{0}{1} = 0\)
- The sum of the product of the roots taken two at a time \(ab + ac + bc = \frac{c}{a} = -1\)
- The product of the roots \(abc = -\frac{d}{a} = -\frac{1}{1} = -1\)

### Step 3: Simplify the expression
We need to find:
\[ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \]
This can be rewritten as:
\[ \frac{(b+1)(c+1) + (c+1)(a+1) + (a+1)(b+1)}{(a+1)(b+1)(c+1)} \]

### Step 4: Expand the numerator
The numerator expands as follows:
\[ (b+1)(c+1) + (c+1)(a+1) + (a+1)(b+1) = (bc + b + c + 1) + (ca + c + a + 1) + (ab + a + b + 1) \]
Combining like terms gives:
\[ ab + ac + bc + 2(a + b + c) + 3 \]
Substituting the values from Vieta's formulas:
- \(ab + ac + bc = -1\)
- \(a + b + c = 0\)

Thus, the numerator becomes:
\[ -1 + 2(0) + 3 = 2 \]

### Step 5: Expand the denominator
The denominator expands as:
\[ (a+1)(b+1)(c+1) = abc + ab + ac + bc + a + b + c + 1 \]
Substituting the values:
- \(abc = -1\)
- \(ab + ac + bc = -1\)
- \(a + b + c = 0\)

Thus, the denominator becomes:
\[ -1 - 1 + 0 + 1 = -1 \]

### Step 6: Combine results
Now we can combine the results:
\[ \frac{2}{-1} = -2 \]

### Final Answer
Thus, the value of \(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\) is:
\[ \boxed{-2} \]