QUESTION IMAGE
Question
express the following fraction in simplest form, only using positive\\(\frac{-3(u^{3}c^{-5})^{-3}}{2u^{-5}c^{9}}\\)
Step1: Apply exponent rule \((a^m)^n = a^{mn}\) to \((u^3c^{-5})^{-3}\)
\[
\]
Step2: Apply exponent rule \(\frac{a^m}{a^n}=a^{m - n}\) for \(u\) and \(c\) terms
For \(u\) terms: \(u^{-9-(-5)} = u^{-9 + 5}=u^{-4}=\frac{1}{u^{4}}\) (but we need positive exponents, so we will handle later)
For \(c\) terms: \(c^{15 - 9}=c^{6}\)
So the expression becomes:
\[
\frac{-3\times\frac{1}{u^{9}}\times c^{15}}{2\times\frac{1}{u^{5}}\times c^{9}}=\frac{-3c^{15}u^{5}}{2c^{9}u^{9}}
\]
Now simplify \(u\) and \(c\) terms:
For \(u\): \(u^{5-9}=u^{-4}=\frac{1}{u^{4}}\) (wait, earlier step had a miscalculation. Let's redo the exponent subtraction correctly. \(\frac{u^{-9}}{u^{-5}}=u^{-9-(-5)} = u^{-4}\) and \(\frac{c^{15}}{c^{9}}=c^{6}\))
So now we have \(\frac{-3}{2}\times u^{-4}\times c^{6}\)
Step3: Convert negative exponent to positive
Using \(a^{-n}=\frac{1}{a^{n}}\), \(u^{-4}=\frac{1}{u^{4}}\), so the expression is \(\frac{-3c^{6}}{2u^{4}}\)
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\(\frac{-3c^{6}}{2u^{4}}\)