Answer:
[tex]3a^4[/tex]
Step-by-step explanation:
What is the cube root of [tex]27a^{12}[/tex]? This is the question.
We can write:
[tex]\sqrt[3]{27a^{12}}[/tex]
We will use the below property to simplify:
[tex]\sqrt[n]{a*b}=\sqrt[n]{a} \sqrt[n]{b}[/tex]
So, we have:
[tex]\sqrt[3]{27a^{12}} =\sqrt[3]{27} \sqrt[3]{a^{12}}[/tex]
We will now use below property to further simplify:
[tex]\sqrt[n]{x} =x^{\frac{1}{n}}[/tex]
Thus, we have:
[tex]\sqrt[3]{27} \sqrt[3]{a^{12}} =3*(a^{12})^{\frac{1}{3}}[/tex]
We know power to the power rule: [tex](a^z)^b=a^{zb}[/tex]
Now, we have:
[tex]3*(a^{12})^{\frac{1}{3}}\\=3*a^{\frac{12}{3}}\\=3a^4[/tex]
This is the correct answer: [tex]3a^4[/tex]
