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

The given equation is:

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

Using the algebraic identity $1 - x^3 = (1-x)(1+x+x^2)$, the left hand side can be written as:

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

Dividing both sides by $(1-x)^{30}$:

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

$\left( \dfrac{1+x+x^2}{(1-x)^2} \right)^{10} = \sum\limits_{r=0}^{10} a_r \left( \dfrac{x}{(1-x)^2} \right)^r$

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

The term inside the bracket on the left hand side can be simplified as:

$\dfrac{1+x+x^2}{(1-x)^2} = \dfrac{1-2x+x^2+3x}{(1-x)^2} = \dfrac{(1-x)^2+3x}{(1-x)^2} = 1 + 3\left(\dfrac{x}{(1-x)^2}\right) = 1 + 3y$

Substituting this into the equation, we get:

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

Using the binomial expansion, $(1 + 3y)^{10} = \sum\limits_{r=0}^{10} {^{10}C_{r}} (3y)^r = \sum\limits_{r=0}^{10} {^{10}C_{r}} 3^r y^r$.

Comparing the coefficients of $y^r$ on both sides, we obtain:

$a_r = {^{10}C_{r}} 3^r$

For $r = 9$ and $r = 10$:

$a_9 = {^{10}C_{9}} 3^9 = 10 \times 3^9$

$a_{10} = {^{10}C_{10}} 3^{10} = 1 \times 3^{10} = 3^{10}$

Therefore, the required value is:

$\dfrac{9a_9}{a_{10}} = \dfrac{9 \times 10 \times 3^9}{3^{10}} = \dfrac{90 \times 3^9}{3 \times 3^9} = \dfrac{90}{3} = 30$

Answer: $30$