# Exercise ID107

Algebra → Roots → Simplifying
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 9] [Number of characters: 0]

Evaluate the following expression:

${243}^{\frac{1}{5}}-{\left(\frac{1}{81}\right)}^{\frac{1}{4}}+{\left(92+7,5\right)}^{0}+{\left({9}^{\frac{7}{8}}:{3}^{\frac{5}{4}}\right)}^{4}$

${243}^{\frac{1}{5}}-{\left(\frac{1}{81}\right)}^{\frac{1}{4}}+{\left(92+7,5\right)}^{0}+{\left({9}^{\frac{7}{8}}:{3}^{\frac{5}{4}}\right)}^{4}$ $=\sqrt[5]{243}-\frac{\sqrt[4]{1}}{\sqrt[4]{81}}+{\left(92+7,5\right)}^{0}+\left({9}^{\frac{7·{4}}{{8}}}:{3}^{\frac{5·{4}}{{4}}}\right)$ $=\sqrt[5]{{3}^{5}}-\frac{\sqrt[4]{1}}{\sqrt[4]{{3}^{4}}}+1+\left({9}^{\frac{7}{2}}:{3}^{5}\right)$ $=3-\frac{1}{3}+1+\frac{{\left(\sqrt{9}\right)}^{7}}{{3}^{5}}$ $=3-\frac{1}{3}+1+\frac{{3}^{7}}{{3}^{5}}$ $=3-\frac{1}{3}+1+{3}^{2}$ $=3-\frac{1}{3}+1+9$ $=\frac{9-1+3+27}{3}$ $=\frac{38}{3}$
${243}^{\frac{1}{5}}-{\left(\frac{1}{81}\right)}^{\frac{1}{4}}+{\left(92+7,5\right)}^{0}+{\left({9}^{\frac{7}{8}}:{3}^{\frac{5}{4}}\right)}^{4}=\frac{40}{3}$

${a}^{\frac{m}{n}}=\sqrt[n]{{a}^{m}}$

${\left({a}^{n}\right)}^{m}={a}^{n·m}$