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)

(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)
Transcript text: $\left(\frac{27 x^{-6} y^{9}}{z^{9}}\right)^{1 / 3}=$ A. $\frac{y^{3}}{x^{2} z^{3}}$ B. $3 y^{3} x^{2} z^{3}$ C. $\frac{3 x^{3}}{y^{2} z^{3}}$ D. $\frac{y^{3}}{3 x^{2} z^{3}}$ E. $\frac{3 y^{3}}{x^{2} z^{3}}$
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Solution

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Solution Steps

Step 1: Simplify the expression inside the parentheses

The given expression is: (27x6y9z9)1/3 \left(\frac{27 x^{-6} y^{9}}{z^{9}}\right)^{1 / 3} First, recognize that 2727 is a perfect cube, since 27=3327 = 3^3. Rewrite the expression as: (33x6y9z9)1/3 \left(\frac{3^3 x^{-6} y^{9}}{z^{9}}\right)^{1 / 3}

Step 2: Apply the exponent 1/31/3 to each term

Using the property (am)n=amn(a^m)^n = a^{m \cdot n}, apply the exponent 1/31/3 to each term inside the parentheses: 3313x613y913z913 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: 31x2y3z3 3^{1} \cdot x^{-2} \cdot y^{3} \cdot z^{-3} This simplifies to: 3x2y3z3 3 \cdot x^{-2} \cdot y^{3} \cdot z^{-3}

Step 3: Rewrite the expression with positive exponents

To make the expression more readable, rewrite x2x^{-2} and z3z^{-3} as 1x2\frac{1}{x^2} and 1z3\frac{1}{z^3}, respectively: 31x2y31z3 3 \cdot \frac{1}{x^2} \cdot y^{3} \cdot \frac{1}{z^3} Combine the terms: 3y3x2z3 \frac{3 y^{3}}{x^{2} z^{3}}

Final Answer

The simplified form of the expression is: 3y3x2z3 \boxed{\frac{3 y^{3}}{x^{2} z^{3}}} The correct choice is E.

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