Coefficient of Cubic Term in Polynomial Difference Function

To find the coefficient of $x^3$ in $h(x) = f(x + 1) - g(x + 2)$, we first examine the difference function $P(x) = f(x) - g(x)$:

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

It is given that $f(x) \neq g(x)$ for all $x \in \mathbb{R}$, which means $P(x) \neq 0$ for all $x \in \mathbb{R}$.

A non-constant polynomial with real coefficients and an odd degree always possesses at least one real root by the Intermediate Value Theorem. For $P(x)$ to have no real roots, its degree cannot be odd.

Therefore:

1. The coefficient of $x^3$ must be zero: $a_3 - b_3 = 0 \implies a_3 = b_3$

2. Furthermore, $a_2 - b_2 \neq 0$ (otherwise, $P(x) = (a_1 - b_1) + 7x$, which has a real root at $x = -\frac{a_1 - b_1}{7}$).

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

1. Coefficient of $x^3$ in $f(x + 1)$: $f(x + 1) = a_1 + 10(x + 1) + a_2(x + 1)^2 + a_3(x + 1)^3 + (x + 1)^4$ The terms contributing to $x^3$ are:

• From $(x + 1)^4$: $\binom{4}{1}x^3(1)^1 = 4x^3$
• From $a_3(x + 1)^3$: $a_3 x^3$

Thus, the coefficient of $x^3$ in $f(x + 1)$ is: $a_3 + 4$

1. Coefficient of $x^3$ in $g(x + 2)$: $g(x + 2) = b_1 + 3(x + 2) + b_2(x + 2)^2 + b_3(x + 2)^3 + (x + 2)^4$ The terms contributing to $x^3$ are:

• From $(x + 2)^4$: $\binom{4}{1}x^3(2)^1 = 8x^3$
• From $b_3(x + 2)^3$: $b_3 x^3$

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

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

Since $a_3 - b_3 = 0$: $\text{Coefficient of } x^3 = 0 - 4 = -4$

Therefore, the correct option is (C).