Question

simplify.
\\(\left(\frac{x^{-3}}{2x^{-1}}\
ight)^3\\)
write your answer using only positive exponents.

Explanation:

Step1: Simplify fraction inside parentheses

Use rule $\frac{a^m}{a^n}=a^{m - n}$. So $\frac{x^{-3}}{2x^{-1}}=\frac{1}{2}x^{-3-(-1)}=\frac{1}{2}x^{-2}$.

Step2: Apply power - of - a - power rule

Use $(ab)^n=a^n b^n$ and $(a^m)^n=a^{mn}$. So $(\frac{1}{2}x^{-2})^3=(\frac{1}{2})^3\times(x^{-2})^3$.
$(\frac{1}{2})^3=\frac{1}{8}$ and $(x^{-2})^3=x^{-6}$.

Step3: Convert to positive exponents

Use $a^{-n}=\frac{1}{a^n}$. So $x^{-6}=\frac{1}{x^6}$. Then $\frac{1}{8}x^{-6}=\frac{1}{8x^6}$.

Answer:

$\frac{1}{8x^6}$