If $$(1 - x^3)^{10} = \displaystyle\sum_{r=0}^{10} a_r x^r (1 - x)^{30 - 2r}$$, then $$\dfrac{9a_9}{a_{10}}$$ is equal to __________.

$$1 - x^3 = (1 - x)(1 + x + x^2)$$

$$(1 - x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{30 - 2r}$$

$$\left[(1 - x)(1 + x + x^2)\right]^{10} = (1 - x)^{10}(1 + x + x^2)^{10}$$

$$(1 - x)^{10}(1 + x + x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{30 - 2r}$$

$$(1 + x + x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{20 - 2r}$$

$$\frac{(1 + x + x^2)^{10}}{(1 - x)^{20}} = \sum_{r=0}^{10} a_r \frac{x^r}{(1 - x)^{2r}}$$

$$\left[ \frac{1 + x + x^2}{(1 - x)^2} \right]^{10} = \sum_{r=0}^{10} a_r \left[ \frac{x}{(1 - x)^2} \right]^r$$

$$\frac{1 + x + x^2}{(1 - x)^2} = \frac{(1 - 2x + x^2) + 3x}{(1 - x)^2} = \frac{(1 - x)^2 + 3x}{(1 - x)^2} = 1 + \frac{3x}{(1 - x)^2} = 1 + 3y$$

Let $$y = \frac{x}{(1 - x)^2}$$

$$(1 + 3y)^{10} = \sum_{r=0}^{10} a_r y^r$$

$$(1 + 3y)^{10} = \sum_{r=0}^{10} \binom{10}{r} (3y)^r = \sum_{r=0}^{10} \binom{10}{r} 3^r y^r$$

$$a_r = \binom{10}{r} 3^r$$

For $$r = 9$$: $$a_9 = \binom{10}{9} 3^9 = 10 \cdot 3^9$$

For $$r = 10$$: $$a_{10} = \binom{10}{10} 3^{10} = 1 \cdot 3^{10}$$

$$\frac{9a_9}{a_{10}} = \frac{9 \cdot (10 \cdot 3^9)}{3^{10}}$$

$$\frac{9a_9}{a_{10}} = \frac{10 \cdot 3^{11}}{3^{10}} = 10 \cdot 3 = 30$$