Consider the polynomial
\[P(x)=x^3+3x^2+6x+10.\]
Let its three roots be $a$, $b$, $c$. Define $Q(x)$ to be the monic cubic polynomial with roots $ab$, $bc$, $ca$. Compute $|Q(1)|$.

[i]Proposed by Nathan Xiong[/i]
1. Given the polynomial \( P(x) = x^3 + 3x^2 + 6x + 10 \), we know its roots are \( a, b, c \). By Vieta's formulas, we have:
   \[
   a + b + c = -3, \quad ab + bc + ca = 6, \quad abc = -10.
   \]

2. Define \( Q(x) \) to be the monic cubic polynomial with roots \( ab, bc, ca \). By the Factor Theorem, we can write:
   \[
   Q(x) = (x - ab)(x - bc)(x - ca).
   \]

3. To find \( |Q(1)| \), we need to evaluate:
   \[
   |Q(1)| = |(1 - ab)(1 - bc)(1 - ca)|.
   \]

4. Expanding the product, we get:
   \[
   (1 - ab)(1 - bc)(1 - ca) = 1 - (ab + bc + ca) + (ab \cdot bc + bc \cdot ca + ca \cdot ab) - (ab \cdot bc \cdot ca).
   \]

5. Substitute the values from Vieta's formulas:
   \[
   ab + bc + ca = 6, \quad ab \cdot bc + bc \cdot ca + ca \cdot ab = (abc)(a + b + c) = (-10)(-3) = 30, \quad ab \cdot bc \cdot ca = (abc)^2 = (-10)^2 = 100.
   \]

6. Therefore, we have:
   \[
   (1 - ab)(1 - bc)(1 - ca) = 1 - 6 + 30 - 100 = -75.
   \]

7. Taking the absolute value, we get:
   \[
   |Q(1)| = |-75| = 75.
   \]

The final answer is \(\boxed{75}\).