The coefficient of $x^{18}$ in the product $(1-x^2)(1-x^3)^9$ is

Question

The coefficient of $x^{18}$ in the product $(1-x^2)(1-x^3)^9$ is

Now obviously for the power to be 18, the expansion of $(1-x^3)^9$ must have a term with the power 16, however if we do that we get $3r=16$ which is obviously not a natural number

however, from this my book has concluded that the coefficient is the coefficient of the term with the power 18 in $(1-x^3)^9$, namely the 6th term, which gives a coefficient of 84.

I'm not sure I understand why this is the correct coefficient

Answer 1

firstly, expend the right side $$(1-x^3)^9$$ to be $$\sum_{k=0}^{6} C_9^k1^{9-k}(-x^3)^k$$ thus $$(1-x^2)(1-x^3)^9 = (1-x^2) * \sum_{k=0}^{6} C_9^k1^{9-k}(-x^3)^k$$ as you can see there are only 2 possibility that term $x^n$ can be $x^{18}$ which would be $$1 * [C_9^k1^{9-k}(-x^3)^k] = c_1 * x^{18}$$ $$-x^{2} * [C_9^k1^{9-k}(-x^3)^k] = c_2 * x^{18}$$ where $c_1$ and $c_2$ are coefficients

neglecting coefficients, to get power 18, it must satisfies that $$x^{3k} = x^{18}$$ or $$x^{2}*x^{3k} = x^{18}$$ the solution is $$ k = 6 $$ or $$ k = \frac{16}{3} $$ since k must be integer so $k = 6$ thus you can solve $c_1$ letting k=6 $$C_9^61^{9-6}(-x^3)^6 = c_1 * x^{18}$$ which is $$84*x^{18} = c_1 x^{18}$$ so $c_1 = 84$ and $c_2$ is impossible

Answer 2

Note that $$ (1-x^2)(1-x^3)^9 = (1-x^3)^9-x^2(1-x^3)^9. $$ The first summand contains the term $(x^3)^6 = x^{18}$. The second does not, as you observe that $(1-x^3)^9$ does not contain $x^{16}$.

Since $$ (1-x^3)^9 = \sum_{k=0}^9 \binom{9}{k} (-x^3)^k = \sum_{k=0}^9 \binom{9}{k} (-1)^k x^{3k}, $$ the coefficient of $x^{18}$ is $$ \binom{9}{6} (-1)^6 = 84. $$