which is the simplified form of $\\left(\\frac{2ab}{a^{-5}b^{2}}\\right)^{-3}$? assume $a\\neq0,b\\neq0$.\n$\…

which is the simplified form of $\\left(\\frac{2ab}{a^{-5}b^{2}}\\right)^{-3}$? assume $a\\neq0,b\\neq0$.\n$\\frac{b^{3}}{8a^{18}}$\n$\\frac{b^{2}}{8a^{45}}$\n$\\frac{a^{6}}{4b}$\n$\\frac{2a^{6}}{b^{5}}$
Answer
Explanation:
Step1: Apply power - of - a - quotient rule
$\left(\frac{2ab}{a^{- 5}b^{2}}\right)^{-3}=\frac{(2ab)^{-3}}{(a^{-5}b^{2})^{-3}}$
Step2: Apply power - of - a - product rule
$(2ab)^{-3}=2^{-3}a^{-3}b^{-3}=\frac{1}{8}a^{-3}b^{-3}$ and $(a^{-5}b^{2})^{-3}=a^{15}b^{-6}$
Step3: Divide the two expressions
$\frac{(2ab)^{-3}}{(a^{-5}b^{2})^{-3}}=\frac{\frac{1}{8}a^{-3}b^{-3}}{a^{15}b^{-6}}=\frac{1}{8}a^{-3 - 15}b^{-3+6}$
Step4: Simplify the exponents
$\frac{1}{8}a^{-18}b^{3}=\frac{b^{3}}{8a^{18}}$
Answer:
$\frac{b^{3}}{8a^{18}}$