Suppose $a$, $b$ and $c$ are integers such that the greatest common divisor of $x^2+ax+b$ and $x^2+bx+c$ is $x+1$ (in the set of polynomials in $x$ with integer coefficients), and the least common multiple of $x^2+ax+b$ and $x^2+bx+c$ $x^3-4x^2+x+6$.  Find $a+b+c$.
1. Given that the greatest common divisor (GCD) of the polynomials \(x^2 + ax + b\) and \(x^2 + bx + c\) is \(x + 1\), we can write:
   \[
   \gcd(x^2 + ax + b, x^2 + bx + c) = x + 1
   \]
   This implies that both polynomials share \(x + 1\) as a factor.

2. Given that the least common multiple (LCM) of the polynomials \(x^2 + ax + b\) and \(x^2 + bx + c\) is \(x^3 - 4x^2 + x + 6\), we can write:
   \[
   \text{lcm}(x^2 + ax + b, x^2 + bx + c) = x^3 - 4x^2 + x + 6
   \]

3. Since the GCD and LCM of two polynomials \(f(x)\) and \(g(x)\) satisfy the relation:
   \[
   f(x) \cdot g(x) = \gcd(f(x), g(x)) \cdot \text{lcm}(f(x), g(x))
   \]
   we can substitute the given polynomials and their GCD and LCM:
   \[
   (x^2 + ax + b)(x^2 + bx + c) = (x + 1)(x^3 - 4x^2 + x + 6)
   \]

4. Factorize the right-hand side:
   \[
   (x + 1)(x^3 - 4x^2 + x + 6) = (x + 1)(x + 1)(x - 2)(x - 3)
   \]
   Simplifying, we get:
   \[
   (x + 1)^2 (x - 2)(x - 3)
   \]

5. Since \(x^2 + ax + b\) and \(x^2 + bx + c\) must be quadratic polynomials, they must be of the form:
   \[
   x^2 + ax + b = (x + 1)(x - 2) = x^2 - x - 2
   \]
   and
   \[
   x^2 + bx + c = (x + 1)(x - 3) = x^2 - 2x - 3
   \]

6. From the above factorizations, we identify the coefficients:
   \[
   a = -1, \quad b = -2, \quad c = -3
   \]

7. Summing these coefficients:
   \[
   a + b + c = -1 - 2 - 3 = -6
   \]

The final answer is \(\boxed{-6}\).