Question Number 99089 by bramlex last updated on 18/Jun/20
$$\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+...}}}}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−...}}}} \\ $$
Answered by MJS last updated on 18/Jun/20
$${a}=\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt[{\mathrm{3}}]{\mathrm{9}+...}} \\ $$$${a}^{\mathrm{3}} =\mathrm{9}+{a} \\ $$$${a}^{\mathrm{3}} −{a}−\mathrm{9}=\mathrm{0} \\ $$$${a}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}} \\ $$$${b}=\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−...}} \\ $$$${b}^{\mathrm{2}} =\mathrm{8}−{b} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${a}−{b}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$
Commented by bemath last updated on 18/Jun/20
$$\mathrm{cardano} \\ $$