求在二项式展开式
$\left(x^2 - \frac{2}{x}\right)^9.$
中不含 $x$ 的项和含有 $\frac{1}{x^3}$ 的项。 由此，求在展开式
$\left(3 - x^3\right)\left(x^2 - \frac{2}{x}\right)^9.$
中不含 $x$ 的项。

1. Write down the general term of the binomial expansion

The general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{2}{x}\right)^r$

2. Simplify the general term to isolate the power of $x$

$T_{r+1} = \binom{9}{r} x^{18-2r} (-2)^r x^{-r}$
$T_{r+1} = \binom{9}{r} (-2)^r x^{18-3r}$

3. Find the term independent of $x$

For the term independent of $x$, the power of $x$ must be $0$:
$18 - 3r = 0$
$3r = 18 \implies r = 6$
Substitute $r = 6$ into the general term:
$T_7 = \binom{9}{6} (-2)^6 x^0$
$T_7 = 84 \times 64 = 5376$

4. Find the $\frac{1}{x^3}$ term

For the $\frac{1}{x^3}$ term, the power of $x$ must be $-3$:
$18 - 3r = -3$
$3r = 21 \implies r = 7$
Substitute $r = 7$ into the general term:
$T_8 = \binom{9}{7} (-2)^7 x^{-3}$
$T_8 = 36 \times (-128) x^{-3} = -4608x^{-3}$

1. Concisely find the required terms from part (i) that will be needed.

The general term of $\left(x^2 - \frac{2}{x}\right)^9$ is:
$T_{r+1} = \binom{9}{r}(x^2)^{9-r}\left(-\frac{2}{x}\right)^r = \binom{9}{r}(-2)^r x^{18-3r}$
For the term independent of $x$, $18 - 3r = 0 \implies r = 6$.
The term is $\binom{9}{6}(-2)^6 = 84 \times 64 = 5376$.
For the $\frac{1}{x^3}$ term, $18 - 3r = -3 \implies r = 7$.
The term is $\binom{9}{7}(-2)^7 x^{-3} = 36 \times (-128) x^{-3} = -4608x^{-3}$.

2. Identify the combinations of terms that produce a term independent of $x$ in the new expansion.

We are expanding $\left(3 - x^3\right)\left(x^2 - \frac{2}{x}\right)^9$.
The term independent of $x$ is formed by:
$3 \times (\text{term independent of } x) + (-x^3) \times \left(\text{term in } \frac{1}{x^3}\right)$

3. Calculate the final term independent of $x$.

Term independent of $x$ $= 3(5376) + (-x^3)(-4608x^{-3})$
$= 16128 + 4608$
$= 20736$