三次方程 $x^3 + ax^2 + bx + c = 0$（其中 $a$, $b$ 和 $c$ 是常数）有根 $3 + \mathrm{i}$ 和 $2$。

1. Identify the condition for complex roots to occur in conjugate pairs

By the Complex Conjugate Root Theorem, if a polynomial has a non-real complex root, its complex conjugate is also a root if and only if all the coefficients of the polynomial are real numbers.
Since the leading coefficient is $1$ (a real number), the required assumption for the root $3 - \mathrm{i}$ to exist alongside $3 + \mathrm{i}$ is that the remaining coefficients $a$, $b$, and $c$ must be real numbers.

1. State all three roots of the cubic equation based on the given assumption.

Given that the assumption in part (i) holds, the coefficients $a$, $b$, and $c$ are real numbers. Therefore, the complex roots must occur in conjugate pairs. The three roots are $2$, $3+\mathrm{i}$, and $3-\mathrm{i}$.

2. Form the quadratic factor corresponding to the complex conjugate roots.

$\begin{aligned} (x - (3+\mathrm{i}))(x - (3-\mathrm{i})) &= ((x-3) - \mathrm{i})((x-3) + \mathrm{i}) \\ &= (x-3)^2 - \mathrm{i}^2 \\ &= (x^2 - 6x + 9) - (-1) \\ &= x^2 - 6x + 10 \end{aligned}$

3. Multiply the quadratic factor by the linear factor to form the cubic polynomial.

$\begin{aligned} (x-2)(x^2 - 6x + 10) &= x(x^2 - 6x + 10) - 2(x^2 - 6x + 10) \\ &= x^3 - 6x^2 + 10x - 2x^2 + 12x - 20 \\ &= x^3 - 8x^2 + 22x - 20 \end{aligned}$

4. Compare coefficients with the given cubic equation to find $a$, $b$, and $c$.

Comparing $x^3 - 8x^2 + 22x - 20 = 0$ with $x^3 + ax^2 + bx + c = 0$, we get:
$a = -8$
$b = 22$
$c = -20$