Question Number 142348 by mathdanisur last updated on 30/May/21
$$\frac{\sqrt[{\mathrm{6}}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{15}}}\:\centerdot\:\sqrt[{\mathrm{3}}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:=\:? \\ $$
Answered by bramlexs22 last updated on 30/May/21
$$\:\frac{\sqrt[{\mathrm{6}\:}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt[{\mathrm{6}\:}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{3}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)}}\:= \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{16}−\mathrm{15}}}\:=\:\mathrm{1} \\ $$
Commented by mathdanisur last updated on 30/May/21
$${thanks}\:{Sir} \\ $$