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

${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $

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

$\sqrt {(3 + \sqrt 5 )} $ is equal to

${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$

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