what is the cube root of $b^{27}$?\n$b^{3}$\n$b^{9}$\n$b^{18}$\n$b^{24}$
what is the cube root of $b^{27}$?\n$b^{3}$\n$b^{9}$\n$b^{18}$\n$b^{24}$
Answer
Explanation:
Step1: Recall cube - root formula
The cube - root of a number $x$ is written as $\sqrt[3]{x}=x^{\frac{1}{3}}$. So, the cube - root of $b^{27}$ is $(b^{27})^{\frac{1}{3}}$.
Step2: Apply power - of - a - power rule
According to the power - of - a - power rule $(a^m)^n=a^{mn}$. Here, $a = b$, $m = 27$, and $n=\frac{1}{3}$. Then $(b^{27})^{\frac{1}{3}}=b^{27\times\frac{1}{3}}$.
Step3: Calculate the exponent
$27\times\frac{1}{3}=9$, so $b^{27\times\frac{1}{3}}=b^{9}$.
Answer:
B. $b^{9}$