Can someone ask me how this problem is solved.
The cubed root of the square root of 27. Why does it equal the square root of 3?
$\sqrt[3]{\sqrt[2]{27}} =\sqrt[2]{\sqrt[3]{27}} = (27^{\frac{1}{3}} ) ^{ \frac{1}{2}} = 3^{\frac{1}{2}} = \sqrt[2]{2} = \sqrt2 $
We are simply manipulating the expression. Since the properties of the exponents are commutative, we can take the cube root of 27 first which is 3 and then take the square root.