`(8a^2b^4)^(1/3)=(8)^(1/3)(a^2)^(1/3)(b^4)^(1/3)`

`=2a^(2/3)b^(4/3)`

In the first line, we used this rule from the last section:

(am)n = amn

That is, we took each item inside the brackets and raised them to the power `1/3`. We can do this because each term is multiplied inside the bracket (if they were added or subtracted, we could not do this).

When we expand this out, the only thing we can do is to find the cube root of `8`, which is `2`, and then just write the a and b parts with fractional powers.

Please support IntMath!