Question

What is the simplest form of 3√27a3b7

Answer 1

Answer:

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

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

Step-by-step 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:

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