[THOUGHT] We need the coefficient of x^3 in h(x)=f(x+1)-g(x+2). The condition is f(x)≠g(x) for all real x, so f-g has no real zeros. Since f and g share x^4 term, f(x)-g(x) is at most cubic; but any real cubic must have a real root unless its leading coefficient is 0. Hence a3=b3. Then compute the x^3 coefficient in h(x) and square it. [/THOUGHT] [PYTHON] import sympy as sp

x=sp.Symbol('x', real=True) a1,a2,a3,b1,b2,b3=sp.symbols('a1 a2 a3 b1 b2 b3', real=True)

f = a1 + 10x + a2x2 + a3*x3 + x4 g = b1 + 3x + b2x2 + b3*x3 + x4 h = sp.expand(f.subs(x, x+1) - g.subs(x, x+2))

coeff_x3 = sp.expand(h).coeff(x,3) print("h(x) =", h) print("coeff of x^3 =", coeff_x3)

apply necessary condition for f(x) != g(x) for all real x

coeff_x3_conditioned = sp.simplify(coeff_x3.subs({a3:b3})) print("coeff of x^3 under a3=b3 =", coeff_x3_conditioned) print("square =", sp.expand(coeff_x3_conditioned**2)) [/PYTHON]

[THOUGHT] From the verified expansion, the coefficient of $x^3$ in $h(x)$ is $a_3-b_3-4$. The condition $f(x)\ne g(x)$ for all real $x$ forces $a_3=b_3$, so the coefficient is $-4$, whose square is $16$.