(a) ${x^{x.{x^{1/3}}}} = {(x\,.\,{x^{1/3}})^x}$$ \Rightarrow $${x^{{x^{1 + {1 \over 3}}}}} = {\left( {{x^{1 + {1 \over 3}}}} \right)^x}$

==> ${x^{{x^{4/3}}}} = {\left( {{x^{4/3}}} \right)^x}$= ${x^{{x^{4/3}}}} = {x^{{4 \over 3}x}} \Rightarrow {x^{4/3}} = {4 \over 3}x$

$\Rightarrow {x^{\frac{4}{3} – 1}} = \frac{4}{3} \Rightarrow {x^{1/3}} = \frac{4}{3}$

 $\therefore x = {\left( {{4 \over 3}} \right)^3} = {{64} \over {27}}$

Also $x=1$ is an obvious solution .