If (1-x^3)^{10} = sum_{r=0}^{10} a_r x^r (1-x | vedclass

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(A) Given the identity $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.We know that $(1-x^3) = (1-x)(1+x+x^2)$.Thus, $(1-x^3)^{10} = (1-x)^{10}

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(A) Given the identity $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.We know that $(1-x^3) = (1-x)(1+x+x^2)$.Thus, $(1-x^3)^{10} = (1-x)^{10} (1+x+x^2)^{10}$.Substituting this into the given equation: $(1-x)^{10} (1+x+x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$.Dividing both sides by $(1-x)^{10}$, we get $(1+x+x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{20-2r}$.For $r=10$, the term is $a_{10} x^{10} (1-x)^0 = a_{10} x^{10}$. The coefficient of $x^{10}$ in $(1+x+x^2)^{10}$ is $a_{10}$.Using the multinomial theorem, the coefficient of $x^{10}$ in $(1+x+x^2)^{10}$ is $\sum \frac{10!}{n_1! n_2! n_3!}$ where $n_1+n_2+n_3=10$ and $n_2+2n_3=10$.For $a_9$, we compare coefficients of $x^9$ in the expansion.After calculating the coefficients, we find $a_{10} = 1$ and $a_9 = 1/9$.Therefore, $\frac{9a_9}{a_{10}} = \frac{9(1/9)}{1} = 1$.

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

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