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. 

It is best to be explicit about the kind of remainder term.
$\def\wi{\subseteq}$
As $x \to 0$:
  $\sin(x) \in x - \frac{1}{6}x^3 + O(x^5)$.
  $\sin(x)^3 \in ( x - \frac{1}{6} x^3 + O(x^5) )^3 \wi x^3 - \frac{1}{2} x^5 + O(x^7)$.
