ABC26GN1885 · Surds and Indices

Given surds are of order 4, 2 and 3 respectively. Their L.C.M. is 12. Changing each to a surd of order 12: $$\sqrt[4]{6}=6^{\frac{1}{4}}=6^{\left(\frac{1}{4}\times\frac{3}{3}\right)}=\left(6^{\frac{3}{12}}\right)=\left(6^{3}\right)^{\frac{1}{12}}=(216)^{\frac{1}{12}}$$ $$\sqrt{2}=2^{\frac{1}{2}}=2^{\left(\frac{1}{2}\times\frac{6}{6}\right)}=\left(2^{\frac{6}{12}}\right)=\left(2^{6}\right)^{\frac{1}{12}}=(64)^{\frac{1}{12}}$$ $$\sqrt[3]{4}=4^{\frac{1}{3}}=4^{\left(\frac{1}{3}\times\frac{4}{4}\right)}=\left(4^{\frac{4}{12}}\right)=\left(4^{4}\right)^{\frac{1}{12}}=(256)^{\frac{1}{12}}$$ Clearly, $(256)^{\frac{1}{12}}>(216)^{\frac{1}{12}}>(64)^{\frac{1}{12}}$. The largest one is $(256)^{\frac{1}{12}}$, i.e. $\sqrt[3]{4}$.