Question 422988
The key here is to get each denominator to the LCD (z+g)(z-g)



{{{(g)/(z+g)-(g)/(z-g)}}}



{{{(g(z-g))/((z+g)(z-g))-(g)/(z-g)}}}



{{{(g(z-g))/((z+g)(z-g))-(g(z+g))/((z+g)(z-g))}}}



{{{(gz-g^2)/((z+g)(z-g))-(gz+g^2)/((z+g)(z-g))}}}



{{{(gz-g^2-(gz+g^2))/((z+g)(z-g))}}}



{{{(gz-g^2-gz-g^2)/((z+g)(z-g))}}}



{{{(-2g^2)/((z+g)(z-g))}}}



So {{{(g)/(z+g)-(g)/(z-g)=(-2g^2)/((z+g)(z-g))}}}



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