what is the cube root of $27a^{12}$?\n- $-3a^{4}$\n- $-3a$\n- $3a$\n- $3a^{4}$

Answer:

D. $3a^{4}$

Explanation:

Step1: Find cube - root of 27

$\sqrt[3]{27}=3$ since $3\times3\times3 = 27$.

Step2: Find cube - root of $a^{12}$

Using the rule $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$, for $n = 3$ and $m = 12$, we have $\sqrt[3]{a^{12}}=a^{\frac{12}{3}}=a^{4}$.

Step3: Combine results

$\sqrt[3]{27a^{12}}=\sqrt[3]{27}\times\sqrt[3]{a^{12}} = 3a^{4}$.