Power series of $\frac{sin(w+z)}{(1-z)(1-w)}$

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I have a question that says to compute all terms of order up to 3 in the power series of the function \begin{equation}\frac{sin(w+z)}{(1-z)(1-w)}\end{equation} At the origin.

I think what we would probably want to do is write the above function as

\begin{equation} \frac{\sin(z)\cos(w)}{(1-z)(1-w)}+\frac{\sin(w)\cos(z)}{(1-z)(1-w)}\end{equation}

And expand one variable at a time, but I dont exactly know how to do that.

If we just consider the first term and expand with respect to $z$, we get

\begin{equation} [z+z^2+\frac{5z^3}{6}+\frac{5z^4}{6}+...]\frac{\cos(w)}{(1-w)}\end{equation}

And then where would you go from here? Would you expand $w$ in a similar way and multiply? And what exactly does it mean by "all terms up to order 3" in a multi variable context? Thanks!