Question Practice

Qn #547
If $x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then ${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =
    ❓ Question: If $x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+... | TANCET Group Studies | Tancet Group Studies