Question

What is the simplest form of RootIndex 3 StartRoot 27 a cubed b Superscript 7 Baseline EndRoot?

Answer 1

Answer:

A

Step-by-step explanation:

If your on e2020 Then A is your answer

Answer 2

Answer: $3ab\sqrt[3]{b^4}$

Step-by-step explanation:

Given the following expression:

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

You need to apply the Product of powers property, which states that:

$(a^m)(a^n)=a^{(m+n)$

Then, you can rewrite the expression as following:

$=\sqrt[3]{27a^3b^4b^3}$

The next step is to descompose 27 into its prime factors:

$27=3*3*3=3^3$

Now you must substitute $3^3$ inside the given root. Then:

$=\sqrt[3]{3^3a^3b^4b^3}$

You need to remember that, according to Radicals properties:

$\sqrt[n]{a^n}=a^{\frac{n}{n}}=a^1=a$

Therefore, the final step is to apply this property in order to finally get the expression is its simplest form. This is:

$=3^{\frac{3}{3}}a^{\frac{3}{3}}b^{\frac{4}{3}}b^{\frac{3}{3}}=3ab^{\frac{4}{3}}b=3ab\sqrt[3]{b^4}$