The problem asks for the square of the coefficient of $x^3$ in $h(x)$, given that $f(x)$ and $g(x)$ satisfy a specific condition for all $x \in \mathbb{R}$. The condition is written as "$f(x) \text{ eq } g(x)$". Given the structure of the polynomials (both monic quartics) and the fact that $f(x)$ and $g(x)$ have different linear coefficients ($10x$ vs $3x$), they cannot be equal. The most standard interpretation in such math problems is an inequality, either $f(x) \ge g(x)$ or $f(x) \le g(x)$ for all $x$.

If $f(x) \ge g(x)$ for all $x$, then the difference polynomial $D(x) = f(x) - g(x)$ must be non-negative for all $x$. $D(x) = (x^4 + a_3 x^3 + a_2 x^2 + 10x + a_1) - (x^4 + b_3 x^3 + b_2 x^2 + 3x + b_1)$ $D(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$.

For a polynomial to be non-negative for all real $x$, its degree must be even, and its leading coefficient must be positive. If the degree were 3 (i.e., $a_3 - b_3 \neq 0$), the range of $D(x)$ would be $(-\infty, \infty)$, meaning it would take negative values. Thus, the coefficient of $x^3$ must be zero. So, $a_3 - b_3 = 0 \implies a_3 = b_3$.

I will use Python to verify the coefficient of $x^3$ in $h(x) = f(x+1) - g(x+2)$ symbolically.

[PYTHON] from sympy import symbols, expand, Poly

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

Define h(x)

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

Expand h(x)

h_expanded = expand(h)

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

coeff_x3 = h_expanded.coeff(x, 3)

print(f"Coefficient of x^3 in h(x): {coeff_x3}") [/PYTHON]

The condition $f(x) \text{ eq } g(x)$ for every $x \in \mathbb{R}$ implies an inequality $f(x) \ge g(x)$ or $f(x) \le g(x)$, as strict equality is impossible due to the differing linear terms ($10x$ vs $3x$). Let $D(x) = f(x) - g(x)$. $D(x) = (x^4 + a_3 x^3 + a_2 x^2 + 10x + a_1) - (x^4 + b_3 x^3 + b_2 x^2 + 3x + b_1)$ $D(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$. For $D(x)$ to be always non-negative or always non-positive, its leading term must have an even degree. Since the coefficient of $x$ is non-zero, $D(x)$ is not a constant. If $a_3 - b_3 \neq 0$, the polynomial is of degree 3, and its range is all real numbers, violating the condition. Thus, the coefficient of $x^3$ in $D(x)$ must be zero: $a_3 - b_3 = 0 \implies a_3 = b_3$.

We want the square of the coefficient of $x^3$ in $h(x) = f(x+1) - g(x+2)$. Using the Python output, the coefficient of $x^3$ in $h(x)$ is $a_3 - b_3 - 4$. Substituting $a_3 = b_3$, the coefficient is $0 - 4 = -4$. The square of the coefficient is $(-4)^2 = 16$.

\boxed{16}