Áp dụng công thức lũy thừa, ta có:

a) $a^{\frac{1}{3}}.\sqrt{a}$

= $a^{\frac{1}{3}}.a^{\frac{1}{2}}$

= $a^{\frac{1}{3}+\frac{1}{2}}=a^{\frac{5}{6}}$

Vậy $a^{\frac{1}{3}}.\sqrt{a}=a^{\frac{5}{6}}$

b) $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}$

= $b^{\frac{1}{2}}.b^{\frac{1}{3}}.b^{\frac{1}{6}}$

= $b^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=b$

Vậy $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}=b$

c) $a^{\frac{4}{3}}:\sqrt[3]{a}$

= $a^{\frac{4}{3}}:a^{\frac{1}{3}}$

= $a^{\frac{4}{3}-\frac{1}{3}}=a$

Vậy $a^{\frac{4}{3}}:\sqrt[3]{a}=a$

d) $\sqrt[3]{b}:b^{\frac{1}{6}}$

= $b^{\frac{1}{3}}:b^{\frac{1}{6}}$

= $b^{\frac{1}{3}-\frac{1}{6}}=b^{\frac{1}{6}}$

Vậy $\sqrt[3]{b}:b^{\frac{1}{6}}=b^{\frac{1}{6}}$