We are given:

$f(x)=a_1+10x+a_2x^2+a_3x^3+x^4$ $g(x)=b_1+3x+b_2x^2+b_3x^3+x^4$ $h(x)=f(x+1)-g(x+2)$

We need the coefficient of $x^3$ in $h(x)$.

Also, $f(x)\ne g(x)$ for every $x\in\mathbb R$.

1. Use the condition $f(x)\ne g(x)$ for every real $x$

Consider

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

because the $x^4$ terms cancel.

So $f(x)-g(x)$ is a cubic polynomial with coefficient of $x$ equal to $7$.

Since $f(x)\ne g(x)$ for every real $x$, the polynomial $f(x)-g(x)$ has no real root.

But any real polynomial of odd degree must have at least one real root. Hence this cubic cannot actually be degree $3$.

So the coefficient of $x^3$ must be $0$:

$(a_3-b_3)=0$

Similarly, if it were degree $1$, it would also have a real root. Since coefficient of $x$ is already $7\ne 0$, the polynomial must be of degree $2$.

Therefore,

$(a_3-b_3)=0$

and the quadratic

$(a_2-b_2)x^2+7x+(a_1-b_1)$

has no real root. Thus its discriminant is negative:

$49-4(a_2-b_2)(a_1-b_1)<0$

This only confirms it is indeed a quadratic with no real roots; in particular,

$a_2-b_2\ne 0$

But for our required coefficient, the key result is:

$a_3=b_3$

2. Find the coefficient of $x^3$ in $f(x+1)$

Expand:

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

We only track the $x^3$ coefficient:

• From $a_1$: contributes $0$
• From $10(x+1)$: contributes $0$
• From $a_2(x+1)^2$: contributes $0$
• From $a_3(x+1)^3$: contributes $a_3$
• From $(x+1)^4 = x^4+4x^3+6x^2+4x+1$: contributes $4$

So coefficient of $x^3$ in $f(x+1)$ is

$a_3+4$

3. Find the coefficient of $x^3$ in $g(x+2)$

Expand:

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

Again track only the $x^3$ coefficient:

• From $b_1$: contributes $0$
• From $3(x+2)$: contributes $0$
• From $b_2(x+2)^2$: contributes $0$
• From $b_3(x+2)^3$: contributes $b_3$
• From $(x+2)^4$: since $(x+2)^4=x^4+8x^3+24x^2+32x+16$ it contributes $8$

So coefficient of $x^3$ in $g(x+2)$ is

$b_3+8$

4. Coefficient of $x^3$ in $h(x)$

Since

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

its $x^3$ coefficient is

$(a_3+4)-(b_3+8)=a_3-b_3-4$

Using $a_3=b_3$,

$a_3-b_3-4=-4$

So the coefficient of $x^3$ in $h(x)$ is

$\boxed{-4}$

5. Match with options

The correct option is:

$\boxed{\text{C: }-4}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They match.