Question

What is the simplest form of 3√27a3b7

Answer 1

`3ab^2\sqrt[3]{b}`

if the problem was
`\sqrt[3]{27a^3b^7}`.

Explanation:

Correct me if I'm wrong by I think you are writing
`\sqrt[3]{27a^3b^7}`.

`\sqrt[3]{27a^3b^7}`

I'm first going to look at this as 3 separate problems and then put it altogether in the end.

Problem 1:
`\sqrt[3]{27}=(3)` since
`(3)^3=27`.

Problem 2:
`\sqrt[3]{a^3}=(a)` since
`(a)^3=a^3`

Problem 3:
`\sqrt[3]{b^7}`. This problem is a little harder because
`b^7`is not a perfect cubes like the others were. But
`b^7` does contain a factor that is a perfect cube. That perfect cube is
`b^6` so rewrite
`b^7` as
`b^6 \cdot b^1` or
`b^6 \cdot b`.

So problem 3 becomes
`\sqrt[3]{b^6 \cdot b}=\sqrt[3]{b^6}\cdot \sqrt[3]{b}=b^2 \cdot \sqrt[3]{b}`. The
`b^2` came from this
`(b^2)^3=b^6`.

Anyways let's put it altogether:

`3ab^2\sqrt[3]{b}`