Question

What is the cube root of 27a^l2?-3a4-3a 3a 3a*​

Answer 1

Answer:

$3a^4$

Step-by-step explanation:

What is the cube root of  $27a^{12}$? This is the question.

We can write:

$\sqrt[3]{27a^{12}}$

We will use the below property to simplify:

$\sqrt[n]{a*b}=\sqrt[n]{a} \sqrt[n]{b}$

So, we have:

$\sqrt[3]{27a^{12}} =\sqrt[3]{27} \sqrt[3]{a^{12}}$

We will now use below property to further simplify:

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

Thus, we have:

$\sqrt[3]{27} \sqrt[3]{a^{12}} =3*(a^{12})^{\frac{1}{3}}$

We know power to the power rule:  $(a^z)^b=a^{zb}$

Now, we have:

$3*(a^{12})^{\frac{1}{3}}\\=3*a^{\frac{12}{3}}\\=3a^4$

This is the correct answer:  $3a^4$