If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $

The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is

${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$

${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $