Showing the simplifying steps of this equality ... Can someone please show me how these are equivalent in steps$$\frac{(h_2^2 - h_3^2 )}{\dfrac{1}{h_3}-\dfrac{1}{h_2}}=h_2 h_3 (h_2+h_3)$$ I thought it simplifies to $$(h_2^2-h_3^2)(h_3-h_2)$$This would be much appreciated, I cant wrap my head around it.

Multiplying the numerator & the denominator by $h_2h_3\ne0,$
$$\frac{(h_2^2 - h_3^2 )}{\dfrac{1}{h_3}-\dfrac{1}{h_2}}= \frac{h_2h_3(h_2-h_3)(h_2+h_3)}{(h_2-h_3)}=h_2h_3(h_2+h_3)$$ assuming $h_2-h_3\ne0$ 
