# Exercise ID275

Algebra → Exponents → Powers
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 9] [Number of characters: 0]
$\left(\frac{z-2}{6z+{\left(z-2\right)}^{2}}+\frac{{\left(z+4\right)}^{2}-12}{{z}^{3}-8}-\frac{1}{z-2}\right):\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}$ $=\left(\frac{z-2}{6z+{z}^{2}-4z+4}+\frac{z+8z+16-12}{\left(z-2\right)\left({z}^{2}+2z+4\right)}-\frac{1}{z-2}\right):\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}$ $=\frac{{\left(z-2\right)}^{2}{+}{{z}}^{2}+8z{+}{4}{-}{{z}}^{2}-2z{-}{4}}{\left(z-2\right)\left({z}^{2}+2z+4\right)}:\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}$ $=\frac{\left({z}^{2}+2z+4\right)}{\left(z-2\right)\left({z}^{2}+2z+4\right)}:\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}$ $=\frac{1}{\left(z-2\right)}:\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}$ $=\frac{1}{\left(z-2\right)}:\frac{{z}^{2}\left(z+2\right)+2\left(z+2\right)}{{z}^{2}\left(z-2\right)+2\left(z-2\right)}$ $=\frac{1}{\left(z-2\right)}:\frac{\left({z}^{2}+2\right)\left(z+2\right)}{\left({z}^{2}+2\right)\left(z-2\right)}$ $=\frac{1}{\left(z-2\right)}·\frac{\left({z}^{2}+2\right)\left(z-2\right)}{\left({z}^{2}+2\right)\left(z+2\right)}$ $=\frac{1}{z+2}$
$\left(\frac{z-2}{6z+{\left(z-2\right)}^{2}}+\frac{{\left(z+4\right)}^{2}-12}{{z}^{3}-8}-\frac{1}{z-2}\right):\frac{{z}^{3}+2{z}^{2}+2z+4}{{z}^{3}-2{z}^{2}+2z-4}=\frac{1}{z+2}$