If (1 - x^3)^10 = Σlimits_r=0^10 a_r x^r (1-x)^30-2r, then 9a_9 a_10 is equal to .

The given equation is:

$$(1 - x^3)^{10} = \sum\limits_{r=0}^{10} a_r x^r (1-x)^{30-2r}$$(1−x3)10=r=0∑10​ar​xr(1−x)30−2r

Using the algebraic identity $1 - x^3 = (1-x)(1+x+x^2)$1−x3=(1−x)(1+x+x2), the left hand side can be written as:

$$(1-x)^{10}(1+x+x^2)^{10} = \sum\limits_{r=0}^{10} a_r x^r (1-x)^{30-2r}$$(1−x)10(1+x+x2)10=r=0∑10​ar​xr(1−x)30−2r

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

$$\dfrac{(1-x)^{10}(1+x+x^2)^{10}}{(1-x)^{30}} = \sum\limits_{r=0}^{10} a_r \dfrac{x^r}{(1-x)^{2r}}$$(1−x)30(1−x)10(1+x+x2)10​=r=0∑10​ar​(1−x)2rxr​

$$\left( \dfrac{1+x+x^2}{(1-x)^2} \right)^{10} = \sum\limits_{r=0}^{10} a_r \left( \dfrac{x}{(1-x)^2} \right)^r$$((1−x)21+x+x2​)10=r=0∑10​ar​((1−x)2x​)r

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

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

$$\dfrac{1+x+x^2}{(1-x)^2} = \dfrac{1-2x+x^2+3x}{(1-x)^2} = \dfrac{(1-x)^2+3x}{(1-x)^2} = 1 + 3\left(\dfrac{x}{(1-x)^2}\right) = 1 + 3y$$(1−x)21+x+x2​=(1−x)21−2x+x2+3x​=(1−x)2(1−x)2+3x​=1+3((1−x)2x​)=1+3y

Substituting this into the equation, we get:

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

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

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

$$a_r = {^{10}C_{r}} 3^r$$ar​=10Cr​3r

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

$$a_9 = {^{10}C_{9}} 3^9 = 10 \times 3^9$$a9​=10C9​39=10×39

$$a_{10} = {^{10}C_{10}} 3^{10} = 1 \times 3^{10} = 3^{10}$$a10​=10C10​310=1×310=310

Therefore, the required value is:

$$\dfrac{9a_9}{a_{10}} = \dfrac{9 \times 10 \times 3^9}{3^{10}} = \dfrac{90 \times 3^9}{3 \times 3^9} = \dfrac{90}{3} = 30$$a10​9a9​​=3109×10×39​=3×3990×39​=390​=30

Answer: $30$30