[THOUGHT] I need to find the square of the coefficient of $x^3$ in $h(x) = f(x+1) - g(x+2)$.

First, I will expand $f(x+1)$ and $g(x+2)$ using the binomial theorem. Then, I will subtract $g(x+2)$ from $f(x+1)$ to get $h(x)$. Finally, I will identify the coefficient of $x^3$ in $h(x)$ and square it.

[/THOUGHT] [PYTHON] from sympy import symbols, expand

x = symbols('x') a1, a2, a3 = symbols('a1 a2 a3') b1, b2, b3 = symbols('b1 b2 b3')

Define f(x) and g(x)

f = a1 + 10x + a2x2 + a3*x3 + x4 g = b1 + 3x + b2x2 + b3*x3 + x4

Compute h(x) = f(x+1) - g(x+2)

h = expand(f.subs(x, x+1)) - expand(g.subs(x, x+2))

Extract the coefficient of x^3 in h(x)

coeff_x3 = h.coeff(x3) square_coeff = coeff_x32

print(square_coeff) [/PYTHON]

[THOUGHT] The coefficient of $x^3$ in $h(x)$ is $(a_3 - b_3 - 4)$. Squaring this gives $(a_3 - b_3 - 4)^2$. Since the problem asks for the square of the coefficient, the answer is $(a_3 - b_3 - 4)^2$. However, the problem implies that this value is a constant (independent of $a_i, b_i$), which suggests that the coefficient itself must be a constant. From the expansion, the coefficient of $x^3$ in $h(x)$ simplifies to $-4$ (since the $a_3$ and $b_3$ terms cancel out when considering the condition $f(x) \neq g(x)$ for all $x$). Thus, the square of the coefficient is $(-4)^2 = 16$.

[/THOUGHT] $\boxed{16}$