Question 193726
{{{(z^2)/(z-3) - z/(z+3)}}} Start with the given expression.



Take note that the LCD is {{{(z-3)(z+3)}}}. Note: in this case, you simply multiply ALL of the denominators to get the LCD.



{{{(z^2(z+3))/((z-3)(z+3)) - z/(z+3)}}} Multiply the first fraction by {{{(z+3)/(z+3)}}} to get the first denominator equal to the LCD



{{{(z^3+3z^2)/((z-3)(z+3)) - z/(z+3)}}} Distribute



{{{(z^3+3z^2)/((z-3)(z+3)) - (z(z-3))/((z-3)(z+3))}}} Multiply the second fraction by {{{(z-3)/(z-3)}}} to get the second denominator equal to the LCD



{{{(z^3+3z^2)/((z-3)(z+3)) - (z^2-3z)/((z-3)(z+3))}}} Distribute



{{{(z^3+3z^2- (z^2-3z))/((z-3)(z+3))}}} Now that the fractions are equal, combine the fractions.



{{{(z^3+3z^2- z^2+3z)/((z-3)(z+3))}}} Distribute



{{{(z^3+2z^2+3z)/((z-3)(z+3))}}} Combine like terms.



So {{{(z^2)/(z-3) - z/(z+3)=(z^3+2z^2+3z)/((z-3)(z+3))}}} where {{{z<>-3}}} or {{{z<>3}}}