The cube root of $9\sqrt 3 + 11\sqrt 2 $ is
$2\sqrt 3 + \sqrt 2 $
$\sqrt 3 + 2\sqrt 2 $
$3\sqrt 3 + \sqrt 2 $
$\sqrt 3 + \sqrt 2 $
If $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$
Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are
The value of $\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $
If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $
${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = $