Answer:
![4x^(3) y^(2) (\sqrt[3]{4 x y})](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bhjol9cnao7aobijtd50hs39jhbfppsf2u.png)
Explanation:
Another complex expression, let's simplify it step by step...
We'll start by re-writing 256 as 4^4
![\sqrt[3]{256 x^(10) y^(7) } = \sqrt[3]{4^(4) x^(10) y^(7) }](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6ohrgzhtoj632lwx1wgwre5wtam1a361pb.png)
Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.
![\sqrt[3]{4^(4) x^(10) y^(7) } = 4 \sqrt[3]{4 x^(10) y^(7) }](https://img.qammunity.org/2020/formulas/mathematics/middle-school/oz8a4uylma3gu3xmp33k1xygvycpsoybci.png)
Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.
![4 \sqrt[3]{4 x^(10) y^(7) } = 4x^(3) \sqrt[3]{4 x y^(7) }](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kozp0oujkvygjkg9tmzape37fo11xr81ob.png)
For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.
![4x^(3) \sqrt[3]{4 x y^(7) } = 4x^(3) y^(2) \sqrt[3]{4 x y}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bdmxoz1aj9q0jhti6mpcu1jwqjx3ogleqp.png)
The answer is then:
![4x^(3) y^(2) \sqrt[3]{4 x y} = 4x^(3) y^(2) (\sqrt[3]{4 x y})](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w05lglpxpig3xb4jfe2h9uubjyuycb9jnf.png)