设有两个三次表达式 $\mathrm{f}(x) = x^3 + (a - 3)x + 2b$ 和 $\mathrm{g}(x) = 3x^3 + x^2 + 5ax + 4b$，其中 $a$ 和 $b$ 为常数。

1. Apply the Factor Theorem to $\mathrm{f}(x)$ to form the first equation.

Since $(x + 3)$ is a factor of $\mathrm{f}(x)$, $\mathrm{f}(-3) = 0$.
$(-3)^3 + (a - 3)(-3) + 2b = 0$
$-27 - 3a + 9 + 2b = 0$
$-3a + 2b = 18 \quad \text{--- (1)}$

2. Apply the Factor Theorem to $\mathrm{g}(x)$ to form the second equation.

Since $(x + 3)$ is a factor of $\mathrm{g}(x)$, $\mathrm{g}(-3) = 0$.
$3(-3)^3 + (-3)^2 + 5a(-3) + 4b = 0$
$3(-27) + 9 - 15a + 4b = 0$
$-81 + 9 - 15a + 4b = 0$
$-15a + 4b = 72 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $a$ and $b$.

Multiply (1) by 2:
$-6a + 4b = 36 \quad \text{--- (3)}$
Subtract (3) from (2):
$(-15a + 4b) - (-6a + 4b) = 72 - 36$
$-9a = 36$
$a = -4 \quad \text{(shown)}$
Substitute $a = -4$ into (1):
$-3(-4) + 2b = 18$
$12 + 2b = 18$
$2b = 6$
$b = 3$