Answer
$2 \sin \ x^3 =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{2x^{6n+3}}{(2n+1)!}$
Work Step by Step
We know that the Maclaurin series for $\sin x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..... $
Replace $x$ with $x^3$ and multiply with $2$ in the above series.
$2\sin x^3=2[x^3-\dfrac{(x^3)^3}{3!}+\dfrac{(x^3)^5}{5!}+.....] =2x^3-\dfrac{x^9}{3}+\dfrac{x^{15}}{60}-.......$
Thus, our required series is:
$2 \sin \ x^3 =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{2x^{6n+3}}{(2n+1)!}$
