Question

NEED HELP WITH THIS PROBLEM!!!!!!

Answer 1

Answer:

  $\displaystyle\frac{\sqrt[4]{3x^2}}{2y}$

Step-by-step explanation:

It can work well to identify 4th powers under the radical, then remove them.

  $\displaystyle\sqrt[4]{\frac{24x^6y}{128x^4y^5}}=\sqrt[4]{\frac{3x^2}{16y^4}}=\sqrt[4]{\frac{3x^2}{(2y)^4}}\\\\=\frac{\sqrt[4]{3x^2}}{2y}$

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The applicable rules of exponents are ...

  1/a^b = a^-b

  (a^b)(a^c) = a^(b+c)

The x-factors simplify as ...

  x^6/x^4 = x^(6-4) = x^2

The y-factors simplify as ...

  y/y^5 = 1/y^(5-1) = 1/y^4

The constant factors simplify in the usual way:

  24/128 = (8·3)/(8·16) = 3/16