Question

which expression is equivalent to \\(\left(\frac{a^{-8}b}{a^{-5}b^3}\
ight)^{-3}\\)? assume \\(a\
eq 0, b\
eq 0\\).\
\\(\bigcirc\\) \\(a^9b^6\\)\
\\(\bigcirc\\) \\(a^9b^{12}\\)\
\\(\bigcirc\\) \\(\frac{1}{a^3b^2}\\)\
\\(\bigcirc\\) \\(\frac{a^{29}}{b^6}\\)

Explanation:

Step1: Simplify the fraction inside the parentheses

Using the rule of exponents $\frac{a^m}{a^n}=a^{m - n}$, for the $a$-terms: $a^{-8 - (-5)} = a^{-3}$, and for the $b$-terms: $b^{1 - 3}=b^{-2}$. So the fraction becomes $\frac{a^{-3}b^{-2}}{1}$ (or just $a^{-3}b^{-2}$).

Step2: Apply the outer exponent

Using the rule $(a^m b^n)^p=a^{mp}b^{np}$, we have $(a^{-3}b^{-2})^{-3}=a^{(-3)\times(-3)}b^{(-2)\times(-3)} = a^{9}b^{6}$.

Answer:

$a^{9}b^{6}$ (corresponding to the first option: $a^9b^6$)