If (1 - x^3)^ 10 = ∑ _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r , then 9a_9 a_...

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$<br><br>Let$y = \dfrac{x}{(1-x)^2}$.<br><br>The term inside the bracket on the left hand side can be simplified as:<br><br>$\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$<br><br>Substituting this into the equation, we get:<br><br>$(1 + 3y)^{10} = \sum\limits_{r=0}^{10} a_r y^r$<br><br>Using the binomial expansion,$(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$.<br><br>Comparing the coefficients of$y^r$on both sides, we obtain:<br><br>$a_r = {^{10}C_{r}} 3^r$<br><br>For$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: $30$