[THOUGHT] Let me analyze this problem step by step.

We have:

• $f(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4$
• $g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4$
• $h(x) = f(x+1) - g(x+2)$

I need to find the coefficient of $x^3$ in $h(x)$.

First, let me compute $f(x+1)$ and $g(x+2)$ symbolically, then find $h(x) = f(x+1) - g(x+2)$.

The key constraint is that $f(x) \neq g(x)$ for every $x \in \mathbb{R}$. This means $f(x) - g(x)$ has no real roots.

Let me compute $f(x) - g(x)$: $f(x) - g(x) = (a_1 - b_1) + (10-3)x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$ $= (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$

For this to have no real roots, we need $a_3 = b_3$ (otherwise it's a cubic which always has at least one real root).

So $a_3 = b_3$, and then: $f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2$

For this quadratic to have no real roots, we need:

• Either $a_2 = b_2$ and $7 = 0$ (impossible), or
• $a_2 \neq b_2$ and the discriminant $< 0$

Wait, if $a_2 = b_2$, then $f(x) - g(x) = (a_1 - b_1) + 7x$, which is linear and has a root unless $7 = 0$, which is false.

So we need $a_2 \neq b_2$ and discriminant $< 0$: $49 - 4(a_2 - b_2)(a_1 - b_1) < 0$

But actually, the coefficient of $x^3$ in $h(x)$ might be determined just by the structure, regardless of specific values. Let me compute it symbolically. [/THOUGHT] [PYTHON] from sympy import symbols, expand, Poly

Define symbols

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

Define f(x) and g(x)

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

Compute f(x+1) and g(x+2)

f_shifted = f.subs(x, x+1) g_shifted = g.subs(x, x+2)

Expand them

f_shifted_expanded = expand(f_shifted) g_shifted_expanded = expand(g_shifted)

print("f(x+1) =", f_shifted_expanded) print("g(x+2) =", g_shifted_expanded)

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

h = expand(f_shifted_expanded - g_shifted_expanded) print("h(x) =", h)

Get the coefficient of x^3

h_poly = Poly(h, x) coeffs = h_poly.all_coeffs() print("Coefficients (from highest to lowest degree):", coeffs)

The coefficient of x^3

coeff_x3 = h_poly.nth(3) print("Coefficient of x^3:", coeff_x3) [/PYTHON]

[THOUGHT] The coefficient of $x^3$ in $h(x)$ is $a_3 - b_3 - 4$.

Now I need to use the constraint that $f(x) \neq g(x)$ for every $x \in \mathbb{R}$.

$f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$

For this polynomial to have no real roots, since it's a polynomial of degree at most 3:

• If $a_3 \neq b_3$, it's a cubic polynomial, which always has at least one real root. So we must have $a_3 = b_3$.

With $a_3 = b_3$: $f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2$

This is a quadratic (or lower degree). For it to have no real roots:

• If $a_2 = b_2$, then $f(x) - g(x) = (a_1 - b_1) + 7x$, which is linear and always has a root (since coefficient of $x$ is 7 ≠ 0).
• So $a_2 \neq b_2$, and the discriminant must be negative: $49 - 4(a_2 - b_2)(a_1 - b_1) < 0$.

Since $a_3 = b_3$, the coefficient of $x^3$ in $h(x)$ is: $a_3 - b_3 - 4 = 0 - 4 = -4$

The square of this coefficient is $(-4)^2 = 16$. [/THOUGHT]

\boxed{16}