Question Number 160831 by MathsFan last updated on 07/Dec/21
$$\mathrm{simplify}\:\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}} \\ $$
Commented by mr W last updated on 07/Dec/21
$$\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{16}+\mathrm{8}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{5}\sqrt{\mathrm{5}}+\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{3}×\mathrm{5}+\mathrm{1}}}{\mathrm{2}} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{5}}\right)^{\mathrm{3}} +\mathrm{3}\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{1}}}{\mathrm{2}} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{3}} }}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\phi={golden}\:{ratio} \\ $$
Commented by MathsFan last updated on 08/Dec/21
$${thank}\:{you}\:{sir} \\ $$$${but}\:{is}\:{there}\:{any}\:{formula}\:{for}\: \\ $$$${questions}\:{like}\:{this}\:{please}? \\ $$
Commented by mr W last updated on 08/Dec/21
$${no}!\:{one}\:{knows}\:{it}\:{or}\:{one}\:{doesn}'{t}\:{know} \\ $$$${it}.\:{there}\:{is}\:{no}\:{general}\:{way}\:{to}\:{solve}\:{it} \\ $$$${through}\:{a}\:{formula}! \\ $$
Commented by MathsFan last updated on 08/Dec/21
$${okay}\:{Sir}. \\ $$$${Thank}\:{you}\:{once}\:{again} \\ $$