Question regarding the Quotient Rule, How does the textbook reach this intermediate step? $$\begin{align*}
\left(\frac{f(x)}{g(x)}\right)’ &= \frac{(x-3)^{1/3}\frac{1}{2}(x+2)^{-1/2}}{(x-3)^{2/3}} - \frac{(x+2)^{1/2}\frac{1}{3}(x-3)^{-2/3}}{(x-3)^{2/3}}\\
&= \frac{(x-3)^{-2/3}(x+2)^{-1/2}}{(x-3)^{2/3}}\cdot\left[\frac{1}{2}(x-3) - \frac{1}{3}(x+2)\right]
\end{align*}$$
Can someone advise how they got to the step after the second equality sign? Step by Step would be most helpful.

$$\dfrac{(x-3)^{1/3}\frac12(x+2)^{-1/2}}{(x-3)^{2/3}}-\dfrac{(x+2)^{1/2}\frac13(x-3)^{-2/3}}{(x-3)^{2/3}}$$
$$=\dfrac{(x-3)\color{blue}{(x-3)^{-2/3}}{\frac12\color{blue}{(x+2)^{-1/2}}}-(x+2)\color{blue}{(x+2)^{-1/2}}\frac13\color{blue}{(x-3)^{-2/3}}}{(x-3)^{2/3}}$$
$$=\dfrac{\color{blue}{(x-3)^{-2/3}(x+2)^{-1/2}}}{(x-3)^{2/3}}\left[\frac12(x-3)-\frac13(x+2)\right].$$
Let me know if you need further details.
