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simplify $(3x^{-2}y^{3}z^{-1})^{2}$

Question

simplify $(3x^{-2}y^{3}z^{-1})^{2}$

Explanation:

Step1: Apply power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). So we apply this to each factor inside the parentheses:
\((3x^{-2}y^{3}z^{-1})^{2}=3^{2}\times(x^{-2})^{2}\times(y^{3})^{2}\times(z^{-1})^{2}\)

Step2: Simplify each term

  • For the constant term: \(3^{2} = 9\)
  • For the \(x\)-term: Using the power of a power rule \((a^m)^n=a^{mn}\), we have \((x^{-2})^{2}=x^{-2\times2}=x^{-4}\)
  • For the \(y\)-term: \((y^{3})^{2}=y^{3\times2}=y^{6}\)
  • For the \(z\)-term: \((z^{-1})^{2}=z^{-1\times2}=z^{-2}\)

Step3: Convert negative exponents to positive (optional, but standard)

Recall that \(a^{-n}=\frac{1}{a^{n}}\), so \(x^{-4}=\frac{1}{x^{4}}\) and \(z^{-2}=\frac{1}{z^{2}}\). Combining all the terms together, we get:
\(9\times\frac{1}{x^{4}}\times y^{6}\times\frac{1}{z^{2}}=\frac{9y^{6}}{x^{4}z^{2}}\)

Answer:

\(\frac{9y^{6}}{x^{4}z^{2}}\)