Simplify square root of 2 over cube root of 2. 2 to the power of one sixth 2 to the power of one third 2 to the power of five sixths 2 to the power of three halves

Answer:2 to the power of one sixthStep-by-step explanation:Assuming you don't already know this, any type of root can be expressed as an exponent. Generally speaking:[tex] \sqrt[n]{x} = {x}^{ \frac{1}{n} } [/tex]So you can rewrite the given fraction as[tex] \frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } } [/tex]and then reduce as you normally would. That is, if the bases of the numerator and denominator are the same, then you can subtract the denominator's exponent from the numerator's exponent like so:[tex] \frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } } = {2}^{ \frac{1}{2} - \frac{1}{3} } [/tex]Since[tex] \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} [/tex]the answer is[tex] {2}^{ \frac{1}{6} } \: or \: \sqrt[6]{2} [/tex]