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Question Number 137582 by SOMEDAVONG last updated on 04/Apr/21

A=(((1/3)((2)^(1/3) −1)((2)^(1/3) +1)^3 ))^(1/3)

$$\mathrm{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Answered by Ñï= last updated on 04/Apr/21

A=(((1/3)((2)^(1/3) −1)((2)^(1/3) +1)^3 ))^(1/3) =(((1/3)((4)^(1/3) −1)((2)^(1/3) +1)^2 ))^(1/3) =(((1/3)((4)^(1/3) −1)((4)^(1/3) +2(2)^(1/3) +1)))^(1/3) =(((1/3)(((16))^(1/3) +2(8)^(1/3) −2(2)^(1/3) +1)))^(1/3) =(((1/3)(2(2)^(1/3) +4−2(2)^(1/3) +1)))^(1/3) =((5/3))^(1/3)

$${A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{16}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{8}}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{4}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$

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