Let denote the set of all real numbers. Let for .Define the functions , , and byf(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4,$$$$g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4,$$$$h(x) = f(x+1) - g(x+2).If for every , find the square of the coefficient of in .\
[THOUGHT] To find the coefficient of (x^3) in (h(x) = f(x+1) - g(x+2)), expand using SymPy. The coefficient will be in terms of (a_3) and (b_3). The condition (f(x) \neq g(x)) for all real (x) implies (f(x) - g(x)) has no real roots. Since (f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3), this cubic must have leading coefficient zero (otherwise it has a real root), so (a_3 = b_3). Thus, the (x^3) coefficient in (h(x)) is independent of (a_i, b_i) and equals -4. Its square is 16. Verify the coefficient with SymPy. [/THOUGHT] [PYTHON] from sympy import symbols, expand
a1, a2, a3, b1, b2, b3, x = symbols('a1 a2 a3 b1 b2 b3 x')
f = a1 + 10x + a2x2 + a3*x3 + x4 g = b1 + 3x + b2x2 + b3*x3 + x4
f_shifted = f.subs(x, x+1) g_shifted = g.subs(x, x+2)
h = expand(f_shifted - g_shifted) coeff_x3 = h.coeff(x**3)
print(coeff_x3) [/PYTHON]
[THOUGHT] The coefficient of (x^3) in (h(x)) is (a_3 - b_3 - 4). Since (f(x) \neq g(x)) for all real (x), (f(x) - g(x)) is a cubic with no real roots, requiring (a_3 - b_3 = 0). Thus, the coefficient is (-4), and its square is 16. [/THOUGHT] \boxed{16}
Explanation
The reasoning begins by expanding (h(x) = f(x+1) - g(x+2)) using SymPy, which yields the coefficient of (x^3) as (a_3 - b_3 - 4). The condition (f(x) \neq g(x)) for all real (x) means (f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3) has no real roots. A cubic polynomial always has at least one real root unless its leading coefficient is zero, so (a_3 - b_3 = 0). Substituting this into the coefficient gives (-4). The square of (-4) is 16, matching the Python output and problem requirements.