Questions: (27 x^(-6) y^9 / z^9)^(1 / 3) = A. y^3 / (x^2 z^3) B. 3 y^3 x^2 z^3 C. 3 x^3 / (y^2 z^3) D. y^3 / (3 x^2 z^3) E. 3 y^3 / (x^2 z^3)

Solution Steps

The given expression is: $$\left(\frac{27 x^{-6} y^{9}}{z^{9}}\right)^{1 / 3}$$ First, recognize that $27$ is a perfect cube, since $27 = 3^3$. Rewrite the expression as: $$\left(\frac{3^3 x^{-6} y^{9}}{z^{9}}\right)^{1 / 3}$$

Using the property $(a^m)^n = a^{m \cdot n}$, apply the exponent $1/3$ to each term inside the parentheses: $$3^{3 \cdot \frac{1}{3}} \cdot x^{-6 \cdot \frac{1}{3}} \cdot y^{9 \cdot \frac{1}{3}} \cdot z^{-9 \cdot \frac{1}{3}}$$ Simplify the exponents: $$3^{1} \cdot x^{-2} \cdot y^{3} \cdot z^{-3}$$ This simplifies to: $$3 \cdot x^{-2} \cdot y^{3} \cdot z^{-3}$$

To make the expression more readable, rewrite $x^{-2}$ and $z^{-3}$ as $\frac{1}{x^2}$ and $\frac{1}{z^3}$, respectively: $$3 \cdot \frac{1}{x^2} \cdot y^{3} \cdot \frac{1}{z^3}$$ Combine the terms: $$\frac{3 y^{3}}{x^{2} z^{3}}$$

Final Answer