computing maclaurin series for $(\sin x)^3$ , order $3$ I have a clarification to ask: 
I want to compute $f(x)=(\sin x)^3$ by maclaurin series, order $n=3$.
I know that: $\sin x=x-\dfrac{x^3}{3!}+R_3(x)$.
So can i say that: $\sin^3x=(\sin x)^3=\left[x-\dfrac{x^3}{3!}+R_3(x)\right]^3=x^3+R_3(x)$?
I'm not sure about my last part: $x^3+R_3(x)$. 
Can someone tell me if that's a true step to do?
Thanks. 

Better use $4\sin^3y=3\sin y-\sin3y$
and $$\sin x=\sum_{r=0}^\infty\dfrac{(-1)^rx^{2r+1}}{(2r+1)!}$$
