Question 545438
{{{(8z)/(z^2-1)}}} + {{{z/(z-1)}}}
Note the z^2 - 1, is the difference of squares, and can be factored to:
{{{(8z)/((z-1)(z+1))}}} + {{{z/(z-1)}}} 
this will be our common denominator and we have
{{{(8z + z(z+1))/((z-1)(z+1))}}} = {{{(8z + z^2+z)/((z-1)(z+1))}}} = {{{(z^2 + 9z )/((z-1)(z+1))}}}