Let $P$ be a cubic monic polynomial with roots $a$, $b$, and $c$.  If $P(1)=91$ and $P(-1)=-121$, compute the maximum possible value of \[\dfrac{ab+bc+ca}{abc+a+b+c}.\]

[i]Proposed by David Altizio[/i]
1. Let \( P(x) \) be a monic cubic polynomial with roots \( \alpha, \beta, \gamma \). Therefore, we can write:
   \[
   P(x) = x^3 + ax^2 + bx + c
   \]
   Given that \( P(1) = 91 \) and \( P(-1) = -121 \), we need to find the maximum value of:
   \[
   \frac{\alpha \beta + \beta \gamma + \gamma \alpha}{\alpha \beta \gamma + \alpha + \beta + \gamma}
   \]

2. By Vieta's formulas, we know:
   \[
   \alpha + \beta + \gamma = -a
   \]
   \[
   \alpha \beta + \beta \gamma + \gamma \alpha = b
   \]
   \[
   \alpha \beta \gamma = -c
   \]

3. Using the given conditions:
   \[
   P(1) = 1^3 + a \cdot 1^2 + b \cdot 1 + c = 1 + a + b + c = 91
   \]
   Therefore:
   \[
   a + b + c = 90
   \]

4. Similarly:
   \[
   P(-1) = (-1)^3 + a \cdot (-1)^2 + b \cdot (-1) + c = -1 + a - b + c = -121
   \]
   Therefore:
   \[
   a - b + c = -120
   \]

5. We now have the system of equations:
   \[
   a + b + c = 90
   \]
   \[
   a - b + c = -120
   \]

6. Subtract the second equation from the first:
   \[
   (a + b + c) - (a - b + c) = 90 - (-120)
   \]
   \[
   2b = 210
   \]
   \[
   b = 105
   \]

7. Add the two equations:
   \[
   (a + b + c) + (a - b + c) = 90 + (-120)
   \]
   \[
   2a + 2c = -30
   \]
   \[
   a + c = -15
   \]

8. We need to find the maximum value of:
   \[
   \frac{\alpha \beta + \beta \gamma + \gamma \alpha}{\alpha \beta \gamma + \alpha + \beta + \gamma} = \frac{b}{-(a + c)}
   \]
   Substituting the values we found:
   \[
   \frac{b}{-(a + c)} = \frac{105}{-(-15)} = \frac{105}{15} = 7
   \]

The final answer is \(\boxed{7}\)