Question

express the following fraction in simplest form, only using positive exponents.\\(dfrac{(3t^{-4}n^{-5})^{-3}}{4n^3}\\)

Explanation:

Step1: Apply power of a product rule

For \((ab)^m = a^m b^m\), so \((3t^{-4}n^{-5})^{-3}=3^{-3}(t^{-4})^{-3}(n^{-5})^{-3}\)

Step2: Apply power of a power rule

For \((a^m)^n = a^{mn}\), so \(3^{-3}(t^{-4})^{-3}(n^{-5})^{-3}=3^{-3}t^{12}n^{15}\)

Step3: Simplify negative exponents

\(3^{-3}=\frac{1}{3^3}=\frac{1}{27}\), so now we have \(\frac{\frac{1}{27}t^{12}n^{15}}{4n^3}\)

Step4: Divide the coefficients and subtract exponents for like bases

For \(n\), using \(\frac{a^m}{a^n}=a^{m - n}\), we get \(\frac{1}{27\times4}t^{12}n^{15 - 3}=\frac{t^{12}n^{12}}{108}\)

Answer:

\(\dfrac{t^{12}n^{12}}{108}\)