Let $$\mathbb{R}$$ denote the set of all real numbers. Let $$a_i, b_i \in \mathbb{R}$$ for $$i \in \{1, 2, 3\}$$. Define the functions $$f: \mathbb{R} \to \mathbb{R}$$, $$g: \mathbb{R} \to \mathbb{R}$$, and $$h: \mathbb{R} \to \mathbb{R}$$ by

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

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

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

If $$f(x) \neq g(x)$$ for every $$x \in \mathbb{R}$$, then the coefficient of $$x^3$$ in $$h(x)$$ is

First isolate the polynomial that encodes the condition “$$f(x)\neq g(x)$$ for every $$x$$”.
Define $$d(x)=f(x)-g(x)\,.$$

Using the given expressions,

$$\begin{aligned} d(x)&=\bigl(a_1+10x+a_2x^2+a_3x^3+x^4\bigr) -\bigl(b_1+3x+b_2x^2+b_3x^3+x^4\bigr)\\ &=\bigl(a_1-b_1\bigr)+\bigl(10-3\bigr)x+\bigl(a_2-b_2\bigr)x^2 +\bigl(a_3-b_3\bigr)x^3. \end{aligned}$$

The $$x^4$$ terms cancel, so $$d(x)$$ is at most cubic: $$d(x)=\alpha+\beta x+\gamma x^2+\delta x^3,$$ where $$\delta=a_3-b_3\,.$$

The statement $$f(x)\neq g(x)\;\forall x\in\mathbb{R}$$ means $$d(x)\neq 0$$ for all real $$x$$. If $$\delta\neq 0$$, then $$d(x)$$ is a genuine cubic. For a real cubic with non-zero leading coefficient,

$$\lim_{x\to\infty}d(x)=\text{sign}(\delta)\,\infty,\qquad \lim_{x\to-\infty}d(x)=-\text{sign}(\delta)\,\infty.$$

The Intermediate Value Theorem then forces at least one real root, contradicting $$d(x)\neq 0$$. Hence the cubic term must vanish:

$$\boxed{a_3-b_3=0}\;\Longrightarrow\;a_3=b_3.$$

Now compute the coefficient of $$x^3$$ in $$h(x)=f(x+1)-g(x+2).$$

First expand $$f(x+1)$$ and $$g(x+2)$$ up to the $$x^3$$ terms.

$$\begin{aligned} f(x+1)&=a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4\\ &=\dots+\bigl(a_3+4\bigr)x^3+\dots\\[6pt] g(x+2)&=b_1+3(x+2)+b_2(x+2)^2+b_3(x+2)^3+(x+2)^4\\ &=\dots+\bigl(b_3+8\bigr)x^3+\dots \end{aligned}$$

(The intermediate “$$\dots$$” terms are not required because only the $$x^3$$ coefficients are needed.)

Subtract to get $$h(x)$$:

Coefficient of $$x^3$$ in $$h(x)$$ $$=\;(a_3+4)-(b_3+8)=a_3-b_3-4.$$

Using $$a_3=b_3$$ established above,

$$a_3-b_3-4=0-4=-4.$$

Therefore the coefficient of $$x^3$$ in $$h(x)$$ is $$\mathbf{-4}$$.

Option C which is: $$-4$$