Question

Answer the problem below

............4
Can (x↑y)2/3 be simplified? yes or no

If yes show how with work if not explain why

Note show your work or ill delete your answer.

Answer 1

It depends on what you mean "simplify". Chances are the goal here is to resolve any exponents that are improper fractions, meaning we don't want exponents larger than 1, like 4/3, for instance, but smaller than 1, like 2/3, is okay.

Now, by the distributive property of exponentiation,

`\left(x^4 y\right)^(2/3) = \left(x^4\right)^(2/3) y^(2/3)`

Since
`(x^a)^b = x^(ab)`, we have

`\left(x^4 y\right)^(2/3) = x^(4\cdot2/3) y^(2/3) = x^(8/3) y^(2/3)`

Convert the improper fraction to a mixed number.

`\frac83 = 3 + \frac23`

Then we can write

`x^(8/3) = x^(3 + 2/3) = x^3 x^(2/3)`

since
`x^(a+b)=x^ax^b`.

Finally, we group together the factors with the same power.

`\left(x^4 y\right)^(2/3) = x^3 x^(2/3) y^(2/3) = \boxed{x^3 (xy)^(2/3)}`