To solve the expression \(\left(\frac{x^3 y^{-2}}{2^2 x^{-5} y^3}\right)^3\), let's first simplify the fraction before applying the exponentiation.
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Simplifying the Fraction: \[ \frac{x^3 y^{-2}}{2^2 x^{-5} y^3} = \frac{x^3}{2^2 x^{-5}} \cdot \frac{y^{-2}}{y^3} \]
Using the quotient of powers rule, \(a^m / a^n = a^{m-n}\):
- For the \(x\) terms: \[ x^{3 - (-5)} = x^{3 + 5} = x^8 \]
- For the \(y\) terms: \[ y^{-2 - 3} = y^{-5} \]
Thus: \[ \frac{x^3 y^{-2}}{2^2 x^{-5} y^3} = \frac{x^8}{2^2 y^5} \]
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Finish Simplifying and Expressing in Standard Form: \[ 2^2 = 4 \] Therefore: \[ \frac{x^8}{4 y^5} \]
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Applying the Cube: Now we calculate: \[ \left(\frac{x^8}{4 y^5}\right)^3 = \frac{x^{8 \cdot 3}}{4^3 y^{5 \cdot 3}} = \frac{x^{24}}{64 y^{15}} \]
So the final result of the expression \(\left(\frac{x^3 y^{-2}}{2^2 x^{-5} y^3}\right)^3\) is: \[ \frac{x^{24}}{64 y^{15}} \]
To respond to the other expressions you mentioned, here's what they become with some key operations:
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StartFraction x Superscript 9 Baseline y Superscript negative 6 Baseline Over 2 Superscript 6 Baseline x Superscript negative 15 Baseline y Superscript 9 Baseline EndFraction: Simplifying gives: \[ \frac{x^{9 - (-15)}}{2^6 y^{-6 - 9}} = \frac{x^{24}}{2^6 y^{-15}} = \frac{x^{24} y^{15}}{64} \]
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\(\left(\frac{1}{2^2 x^2}\right)^3\): \[ = \frac{1^3}{(2^2)^3 (x^2)^3} = \frac{1}{8 x^6} \]
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\(\left(\frac{1}{x^8}\right)^3 = \frac{1}{x^{24}}\)
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\(\left(\frac{1}{2^2 x^{15} y^6}\right)^3 = \frac{1}{(2^2)^3 (x^{15})^3 (y^6)^3} = \frac{1}{64 x^{45} y^{18}}\)
Each of these follows similar quotient and power rules as shown above!