How do I go about finding the answer to this? I have tried two methods, using the known taylor series for $\sin (x)$ and $\frac{1}{1-r}$ and then transform and add those, and I have also tried by finding the first $6$ derivatives of the function. However, I am confused since after the first derivative, all of the derivatives at $0$ are $0$, which leaves me with only the first term?
Any help would be much appreciated! Thank you.
On
To deal with the $x^3$, observe that you can apply the geometric series trick
$$\frac{1}{8-x^3}=\frac{1}{8}\frac{1}{1-\frac{x^3}{8}}=\frac{1}{8}\Big(1+\frac{x^3}{8}+\frac{x^6}{64}+\frac{x^9}{512}+\dots\Big)$$
and manipulate the Maclaurin Expansion (a Taylor series expansion centered at $x=0$). The Maclaurin Expansion of $\sin(x)$ is
$$\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$
so to find the Maclaurin Expansion of $\sin(x^3)$ observe that
$$\sin(x^3)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(x^3)^{2k+1}=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-\frac{x^{21}}{7!}+\dots$$
Apply the same strategy to $x^2\sin(x^3)$ and then add this to $\frac{1}{8-x^3}$.
Hint
$$\sin x=x-{x^3\over 6}+\cdots$$and $${1\over {8-x}}={1\over 8}\left(1+{x\over 8}+{x^2\over 64}+{x^3\over 512}+\cdots\right)$$and calculate the terms consisting $x^\alpha$ in the Taylor expansion of $f(x)$ where $\alpha\in\{0,1,2,3,4,5\}$