Question Number 156966 by Armindo last updated on 17/Oct/21
Commented by Armindo last updated on 17/Oct/21
Helô, I need help...
Answered by Rasheed.Sindhi last updated on 18/Oct/21
$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:−\mathrm{1}}=\frac{\left(\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\right)^{\mathrm{2}} −\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:−\mathrm{1}}=\frac{\cancel{\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}\:−\mathrm{1}\right)}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\mathrm{1}\right)}{\:\cancel{\sqrt[{\mathrm{3}}]{\mathrm{3}}\:−\mathrm{1}}} \\ $$$$=\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\mathrm{1} \\ $$
Answered by depressiveshrek last updated on 18/Oct/21
$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}×\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}} \\ $$$$\: \\ $$$$\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}\right)}{\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}\right)} \\ $$$$\: \\ $$$$\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}\right)−\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}^{\mathrm{3}} }−\mathrm{1}^{\mathrm{3}} } \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}\ast\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{9}\ast\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}−\sqrt[{\mathrm{3}}]{\mathrm{9}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\mathrm{3}−\mathrm{1}} \\ $$$$\:\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{81}}+\sqrt[{\mathrm{3}}]{\mathrm{27}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}^{\mathrm{3}} \ast\mathrm{3}}+\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}^{\mathrm{3}} }\ast\sqrt[{\mathrm{3}}]{\mathrm{3}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{2}}{\mathrm{2}} \\ $$$$\: \\ $$$$\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{2}}{\mathrm{2}} \\ $$$$\: \\ $$$$\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{2}}{\mathrm{2}} \\ $$$$\: \\ $$$$\frac{\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{1} \\ $$$$\:\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}−\sqrt[{\mathrm{4}}]{\mathrm{2}}} \\ $$$$\: \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}−\sqrt[{\mathrm{4}}]{\mathrm{2}}}×\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\left(\sqrt[{\mathrm{4}}]{\mathrm{5}}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} }−\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} }} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\frac{\left(\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)}{\left(\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\sqrt{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{5}}\sqrt{\mathrm{2}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\sqrt{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}^{\mathrm{2}} }−\sqrt{\mathrm{2}^{\mathrm{2}} }} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{5}}\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} }}{\mathrm{5}−\mathrm{2}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}\ast\mathrm{5}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{5}\ast\mathrm{2}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{2}\ast\mathrm{5}^{\mathrm{2}} }+\sqrt[{\mathrm{4}}]{\mathrm{2}\ast\mathrm{2}^{\mathrm{2}} }}{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{3}} }+\sqrt[{\mathrm{4}}]{\mathrm{5}\ast\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{2}\ast\mathrm{25}}+\sqrt[{\mathrm{4}}]{\mathrm{2}\ast\mathrm{4}}}{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{125}}+\sqrt[{\mathrm{4}}]{\mathrm{20}}+\sqrt[{\mathrm{4}}]{\mathrm{50}}+\sqrt[{\mathrm{4}}]{\mathrm{8}}}{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} \ast\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} \ast\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} \ast\mathrm{2}}+\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} \ast\mathrm{2}}}{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} }\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} }\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt[{\mathrm{4}}]{\mathrm{5}^{\mathrm{2}} }\sqrt[{\mathrm{4}}]{\mathrm{2}}+\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{2}} }\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\sqrt{\mathrm{5}}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{5}}\sqrt[{\mathrm{4}}]{\mathrm{2}}+\sqrt{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\mathrm{3}} \\ $$$$\: \\ $$
Commented by Armindo last updated on 18/Oct/21
thanks...I will check the procedure!
do you have a book, when I find that solve or This procedure?