The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
$\sqrt 3 + \sqrt 7 $
$2\sqrt 3 + \sqrt 7 $
$\sqrt 3 + 2\sqrt 7 $
None of these
${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
$\sqrt {(3 + \sqrt 5 )} $ is equal to
If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/x) = } $
If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $