Answer:
[tex]3ab^2\sqrt[3]{b}[/tex]
if the problem was [tex]\sqrt[3]{27a^3b^7}[/tex].
Step-by-step explanation:
Correct me if I'm wrong by I think you are writing [tex]\sqrt[3]{27a^3b^7}[/tex].
[tex]\sqrt[3]{27a^3b^7}[/tex]
I'm first going to look at this as 3 separate problems and then put it altogether in the end.
Problem 1: [tex]\sqrt[3]{27}=(3)[/tex] since [tex](3)^3=27[/tex].
Problem 2:[tex]\sqrt[3]{a^3}=(a)[/tex] since [tex](a)^3=a^3[/tex]
Problem 3: [tex]\sqrt[3]{b^7}[/tex]. This problem is a little harder because [tex]b^7[/tex]is not a perfect cubes like the others were. But [tex]b^7[/tex] does contain a factor that is a perfect cube. That perfect cube is [tex]b^6[/tex] so rewrite [tex]b^7[/tex] as [tex]b^6 \cdot b^1[/tex] or [tex]b^6 \cdot b[/tex].
So problem 3 becomes [tex]\sqrt[3]{b^6 \cdot b}=\sqrt[3]{b^6}\cdot \sqrt[3]{b}=b^2 \cdot \sqrt[3]{b}[/tex]. The [tex]b^2[/tex] came from this [tex](b^2)^3=b^6[/tex].
Anyways let's put it altogether:
[tex]3ab^2\sqrt[3]{b}[/tex]
