Suppose `a ,b ,c in I` such that the greatest common divisor for `x^2+a x+b` and `x^2+bx+c` is `(x+1)` and the least common multiple of `x^2+a x+b` and `x^2+b x+c` is `(x^3-4x^2+x+6)`. Then the value of `|a+b+c|` is equal to ___________.

Explanation

To solve the problem step by step, we will follow the reasoning from the video transcript and derive the values of \(a\), \(b\), and \(c\) systematically.

### Step 1: Understanding the GCD
We are given that the greatest common divisor (GCD) of the polynomials \(x^2 + ax + b\) and \(x^2 + bx + c\) is \(x + 1\). This implies that both polynomials can be expressed as:
\[ x^2 + ax + b = (x + 1)(x + d) \quad \text{for some integer } d \]
\[ x^2 + bx + c = (x + 1)(x + e) \quad \text{for some integer } e \]

### Step 2: Expanding the Polynomials
Expanding both expressions, we get:
1. \(x^2 + ax + b = x^2 + (1 + d)x + d\)
- From this, we can equate coefficients:
- \(a = 1 + d\)
- \(b = d\)

2. \(x^2 + bx + c = x^2 + (1 + e)x + e\)
- From this, we can equate coefficients:
- \(b = 1 + e\)
- \(c = e\)

### Step 3: Relating \(a\), \(b\), and \(c\)
Now we have:
- From the first polynomial: \(a = 1 + d\) and \(b = d\)
- From the second polynomial: \(b = 1 + e\) and \(c = e\)

### Step 4: Expressing \(d\) and \(e\)
From \(b = d\) and \(b = 1 + e\), we can set these equal:
\[ d = 1 + e \implies e = d - 1 \]
Substituting \(e\) into \(c\):
\[ c = e = d - 1 \]

### Step 5: Substituting \(d\) into \(a\), \(b\), and \(c\)
Now substituting \(d\) into the equations:
- \(b = d\)
- \(a = 1 + d\)
- \(c = d - 1\)

### Step 6: Finding \(a\), \(b\), and \(c\)
From \(b = d\), we can express \(a\) and \(c\) in terms of \(b\):
- \(a = 1 + b\)
- \(c = b - 1\)

### Step 7: Using the LCM
We are also given that the least common multiple (LCM) of the two polynomials is:
\[ x^3 - 4x^2 + x + 6 \]
Using the relationship between GCD and LCM:
\[ \text{GCD} \cdot \text{LCM} = (x^2 + ax + b)(x^2 + bx + c) \]
Substituting the GCD and LCM:
\[ (x + 1)(x^3 - 4x^2 + x + 6) = (x^2 + ax + b)(x^2 + bx + c) \]

### Step 8: Expanding and Comparing Coefficients
Expanding the left-hand side:
\[ (x + 1)(x^3 - 4x^2 + x + 6) = x^4 - 4x^3 + x^2 + 6x + x^3 - 4x^2 + x + 6 = x^4 - 3x^3 - 3x^2 + 7x + 6 \]
Now, expanding the right-hand side:
\[ (x^2 + ax + b)(x^2 + bx + c) = x^4 + (a + b)x^3 + (ab + c + a)x^2 + (ac + b^2)x + bc \]
Equating coefficients gives us:
1. \(a + b = -3\)
2. \(ab + c + a = -3\)
3. \(ac + b^2 = 7\)
4. \(bc = 6\)

### Step 9: Solving the Equations
From \(a + b = -3\), we can substitute \(a = 1 + b\):
\[ (1 + b) + b = -3 \implies 1 + 2b = -3 \implies 2b = -4 \implies b = -2 \]
Substituting \(b = -2\) into \(a\):
\[ a = 1 + (-2) = -1 \]
Now substituting \(b = -2\) into \(c\):
\[ c = b - 1 = -2 - 1 = -3 \]

### Step 10: Final Calculation
Now we have:
- \(a = -1\)
- \(b = -2\)
- \(c = -3\)

Calculating \(|a + b + c|\):
\[ |a + b + c| = |-1 - 2 - 3| = |-6| = 6 \]

### Final Answer
The value of \(|a + b + c|\) is \(6\).