通过考虑 $\left(x^2 - \frac{1}{3x}\right)^9$ 的二项式展开式中的通项，解释为什么 $x$ 的次数总是能被 3 整除。 求 $\left(x^2 - \frac{1}{3x}\right)^9 (1 + 3x^3)^2$ 展开式中的常数项。

1. Write down the general term of the binomial expansion

The general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r$

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

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

3. Explain why the power of $x$ is divisible by 3

The power of $x$ is $18 - 3r = 3(6 - r)$.
Since $r$ is an integer ($0 \leqslant r \leqslant 9$), $6 - r$ is an integer.
Therefore, $3(6 - r)$ is a multiple of 3, meaning the powers of $x$ are always divisible by 3.

1. Expand the second bracket

$(1 + 3x^3)^2 = 1 + 6x^3 + 9x^6$

2. Identify the general term of the first bracket

The general term of $\left(x^2 - \frac{1}{3x}\right)^9$ is:
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r = \binom{9}{r} \left(-\frac{1}{3}\right)^r x^{18-3r}$

3. Determine the values of $r$ needed to form terms independent of $x$

To get a term independent of $x$ when multiplied by $(1 + 6x^3 + 9x^6)$, we need the powers of $x$ from the first bracket to be $0$, $-3$, and $-6$.
For $x^0$: $18 - 3r = 0 \implies r = 6$
For $x^{-3}$: $18 - 3r = -3 \implies r = 7$
For $x^{-6}$: $18 - 3r = -6 \implies r = 8$

4. Calculate the coefficients for these values of $r$

For $r = 6$: $\binom{9}{6} \left(-\frac{1}{3}\right)^6 = 84 \times \frac{1}{729} = \frac{28}{243}$
For $r = 7$: $\binom{9}{7} \left(-\frac{1}{3}\right)^7 = 36 \times \left(-\frac{1}{2187}\right) = -\frac{4}{243}$
For $r = 8$: $\binom{9}{8} \left(-\frac{1}{3}\right)^8 = 9 \times \frac{1}{6561} = \frac{1}{729}$

5. Calculate the final term independent of $x$

Term independent of $x$:
$= \left(\frac{28}{243}\right)(1) + \left(-\frac{4}{243}\right)(6) + \left(\frac{1}{729}\right)(9)$
$= \frac{28}{243} - \frac{24}{243} + \frac{3}{243}$
$= \frac{7}{243}$