Generating functions - difficulty with question

One way to simplify the computation would be to take out the $\frac{1}{1-x^3}$, so that you only have to do partial fractions on $\frac{1}{(1-x^2)(1-x)}$. Then you get \begin{align*} \frac{1}{(1-x^2)(1-x)} &= \frac{1}{4(1+x)} + \frac{1}{4(1-x)} + \frac{1}{2(1-x)^2} \\ &= \frac{1}{4} \sum_{i=0}^\infty (-x)^i + \frac{1}{4} \sum_{i=0}^\infty x^i + \frac{1}{2} \sum_{i=0}^\infty (i+1)x^i \\ &= \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right] \end{align*} Then, $$ g(x) = (1 + x^3 + x^6 + x^9 + \cdots) \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right] $$ Your answer is the coefficient of $x^{10}$ in $g(x)$, which is, using the above, \begin{align*} \quad \frac{(-1)^{10} + 1 + 2(10 + 1)}{4} &+ \frac{(-1)^{7} + 1 + 2(7 + 1)}{4} \\ + \frac{(-1)^{4} + 1 + 2(4 + 1)}{4} &+ \frac{(-1)^{1} + 1 + 2(1 + 1)}{4} \\ &= \frac{24}{4} + \frac{16}{4} + \frac{12}{4} + \frac{4}{4} \\ &= 6 + 4 + 3 + 1 \\ &= 14. \end{align*}