$\frac{1}{\sqrt{9}-\sqrt{8}}$  is equal to

The number obtained on rationalizing the denominator of  $\frac{3}{\sqrt{3}}$  is

 $64^{\frac{1}{3}}$  is equal to

 $2^{\frac{1}{3}}\times\ 2^{\frac{1}{6}}$  is equal to

 $a^{-1}$  is equal to 

 $\left(3^{\frac{1}{2}}\right)^{\frac{1}{3}}$  is equal to