Ratio of Coefficients in Binomial Expansion

To find the value of $\frac{9a_9}{a_{10}}$, we start with the given identity:

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

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

$(1-x^3)^{10} = (1-x)^{10} (1+x+x^2)^{10}$

Substituting this into the given identity gives:

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

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

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

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

Observe that the expression inside the brackets on the LHS can be rewritten as:

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

Let $y = \frac{x}{(1-x)^2}$. Substituting $y$ into the equation yields:

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

Expanding $(1+3y)^{10}$ using the binomial theorem gives:

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

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

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

Now, calculating $a_9$ and $a_{10}$:

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

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

Substitute $a_9$ and $a_{10}$ into the required expression:

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