If ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ and ${b^2} = ac$ then $x + z = $
$y$
$2y$
$2xyz$
None of these
${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $
$\root 4 \of {(17 + 12\sqrt 2 )} = $
Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution
Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $