Solve z^2-9/9z+27div3z^2-18z+27/27z+81

$$\frac { z ^ { 2 } - 9 } { 9 z + 27 } \div \frac { 3 z ^ { 2 } - 18 z + 27 } { 27 z + 81 }$$

Evaluate

$$\frac{z+3}{z-3}$$

Short Solution Steps

Divide \frac{z^{2}-9}{9z+27} by \frac{3z^{2}-18z+27}{27z+81} by multiplying \frac{z^{2}-9}{9z+27} by the reciprocal of \frac{3z^{2}-18z+27}{27z+81}.

$$\frac{\left(z^{2}-9\right)\left(27z+81\right)}{\left(9z+27\right)\left(3z^{2}-18z+27\right)}$$

Factor the expressions that are not already factored.

$$\frac{27\left(z-3\right)\left(z+3\right)^{2}}{3\times 9\left(z+3\right)\left(z-3\right)^{2}}$$

Cancel out 3\times 9\left(z-3\right)\left(z+3\right) in both numerator and denominator.

$$\frac{z+3}{z-3}$$

Expand

$$\frac{z+3}{z-3}$$

Short Solution Steps

Divide \frac{z^{2}-9}{9z+27} by \frac{3z^{2}-18z+27}{27z+81} by multiplying \frac{z^{2}-9}{9z+27} by the reciprocal of \frac{3z^{2}-18z+27}{27z+81}.

$$\frac{\left(z^{2}-9\right)\left(27z+81\right)}{\left(9z+27\right)\left(3z^{2}-18z+27\right)}$$

Factor the expressions that are not already factored.

$$\frac{27\left(z-3\right)\left(z+3\right)^{2}}{3\times 9\left(z+3\right)\left(z-3\right)^{2}}$$

Cancel out 3\times 9\left(z-3\right)\left(z+3\right) in both numerator and denominator.

$$\frac{z+3}{z-3}$$