Since \((a, m) = 1\) and \((b, m) = 1\):
$$ax+my=1$$$$bw+mz=1$$
Rearrange:
$$ax=1-my$$$$bw=1-mz$$
Multiply them:
$$\begin{align} (ax)(bw) & = (1-my)(1-mz) \\ & =(1-my-mz+m^2yz) \end{align}$$
Simplify:
$$ ab(xw)=1+m(-y-z+myz) $$$$ 1=ab(xw)+m(y+z-myz) $$
Since \(1 = ab(j)+m(k)\), then \((ab, m) = 1\). As a corollary, we can also claim that \((a,b)=1\), then \((a^c,b)=1\), where \(c\) is any positive integer.