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Question Number 142348 by mathdanisur last updated on 30/May/21

(((4−(√(15))))^(1/6) /( (√(4−(√(15)))) ∙ ((4+(√(15))))^(1/3) )) = ?

$$\frac{\sqrt[{\mathrm{6}}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{15}}}\:\centerdot\:\sqrt[{\mathrm{3}}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:=\:? \\ $$

Answered by bramlexs22 last updated on 30/May/21

 (((4−(√(15))))^(1/(6 )) /( (((4−(√(15)))^3 ))^(1/(6 )) .((4+(√(15))))^(1/(3 )) )) =    (1/( (((4−(√(15)))^2 ))^(1/6) .((4+(√(15))))^(1/(3 )) )) =   (1/( (((4−(√(15)))(4+(√(15)))))^(1/3) )) =   (1/( ((16−15))^(1/3) )) = 1

$$\:\frac{\sqrt[{\mathrm{6}\:}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt[{\mathrm{6}\:}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{3}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)}}\:= \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{16}−\mathrm{15}}}\:=\:\mathrm{1} \\ $$

Commented by mathdanisur last updated on 30/May/21

thanks Sir

$${thanks}\:{Sir} \\ $$

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