Let $ℝ$ denote the set of all real numbers. Let $a_i,b_i\in ℝ\text{for}i\in \{1,2,3\}$.

Define the functions $f:ℝ\to ℝ$, $g:ℝ\to ℝ$, and $h:ℝ\to ℝ$ by

$f\left(x\right)=a_1+10x+a_2{x}^2+a_3{x}^3+x^4$

$g\left(x\right)=b_1+3x+b_2{x}^2+b_3{x}^3+x^4$

$h\left(x\right)=f(x+1)-g(x+2)$

If $f\left(x\right)=g\left(x\right)$ for every $x\in ℝ$, then the coefficient of $x^3$ in $h\left(x\right)$ is

We are given the functions $f,g,h:ℝ\to ℝ$ defined by:

$f\left(x\right)=x^4+a_3{x}^3+a_2{x}^2+10x+a_1$

$g\left(x\right)=x^4+b_3{x}^3+b_2{x}^2+3x+b_1$

and the function $h\left(x\right)$ is given by:

$h\left(x\right)=f(x+1)-g(x+2)$

We are told that $f\left(x\right)=g\left(x\right)$ for all $x\in ℝ$. For these two polynomial functions to be equal for all real numbers, their corresponding coefficients must be equal. In particular, the coefficients of the $x^3$ terms must be equal:
$a_3=b_3$

Now, let's determine the coefficient of $x^3$ in both $f(x+1)$ and $g(x+2)$.

Step 1: Coefficient of $x^3$ in $f(x+1)$
We substitute $x+1$ into $f\left(x\right)$:
$f(x+1)={(x+1)}^4+a_3{(x+1)}^3+a_2{(x+1)}^2+10(x+1)+a_1$

Using the binomial expansion, the only terms that can produce $x^3$ are:
1. From ${(x+1)}^4$: The $x^3$ term is $\left(\genfrac{}{}{0ex}{}{4}{1}\right)x^3\left(1\right)=4x^3$.
2. From $a_3{(x+1)}^3$: The $x^3$ term is $a_3{x}^3$.
Therefore, the coefficient of $x^3$ in $f(x+1)$ is:
$4+a_3$

Step 2: Coefficient of $x^3$ in $g(x+2)$
We substitute $x+2$ into $g\left(x\right)$:
$g(x+2)={(x+2)}^4+b_3{(x+2)}^3+b_2{(x+2)}^2+3(x+2)+b_1$

Using the binomial expansion, the only terms that can produce $x^3$ are:
1. From ${(x+2)}^4$: The $x^3$ term is $\left(\genfrac{}{}{0ex}{}{4}{1}\right)x^3\left(2\right)=8x^3$.
2. From : The $x^3$ term is $b_3{x}^3$.
Therefore, the coefficient of $x^3$ in $g(x+2)$ is:
$8+b_3$

Step 3: Coefficient of $x^3$ in $h\left(x\right)$
Using the definition of $h\left(x\right)$, the coefficient of $x^3$ is:
$(4+a_3)-(8+b_3)=-4+a_3-b_3$

Since we established that $a_3=b_3$, we have:
$a_3-b_3=0$

Substituting this back into our expression for the coefficient of $x^3$ in $h\left(x\right)$ yields:
$-4+0=-4$

Therefore, the coefficient of $x^3$ in $h\left(x\right)$ is -4.