To solve for the coefficient of $ x^3 $ in the function $ h(x) = f(x + 1) - g(x + 2) $, we start by expanding both $ f(x + 1) $ and $ g(x + 2) $.

First, expand $ f(x + 1) $:

$ f(x+1) = a_1 + 10(x+1) + a_2 (x+1)^2 + a_3 (x+1)^3 + (x+1)^4 $

From this expansion, the coefficient of $ x^3 $ in $ f(x+1) $ is $ a_3 + 4 $.

Next, expand $ g(x + 2) $:

$ g(x+2) = b_1 + 3(x+2) + b_2 (x+2)^2 + b_3 (x+2)^3 + (x+2)^4 $

Here, the coefficient of $ x^3 $ in $ g(x+2) $ is $ b_3 + 8 $.

Now, calculate the coefficient of $ x^3 $ in $ h(x) = f(x+1) - g(x+2) $:

$ \text{Coefficient of } x^3 \text{ in } h(x) = (a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4 $

Given that $ f(x) \neq g(x) $ for every $ x \in \mathbb{R} $, it follows that $ f(x) - g(x) \neq 0 $ for any $ x $. This means the equation:

$ (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3 = 0 $

cannot have any real roots. For this polynomial to have no real roots, the degree of the polynomial must be less than or equal to zero, implying that the coefficients of the highest degree terms must equate to zero. Thus, $ a_3 - b_3 = 0 $.

Substituting $ a_3 - b_3 = 0 $ into our earlier expression for the coefficient of $ x^3 $ in $ h(x) $ gives:

$ a_3 - b_3 - 4 = -4 $

Thus, the coefficient of $ x^3 $ in $ h(x) $ is $-4$.