To simplify the expression \((4x^4 y^{-4})^3/(x^7y^3)^2\) using properties of integer exponents, we can proceed step by step.
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Simplify the numerator: \[ (4x^4 y^{-4})^3 = 4^3 (x^4)^3 (y^{-4})^3 \] Now calculate each part: \[ 4^3 = 64, \quad (x^4)^3 = x^{12}, \quad (y^{-4})^3 = y^{-12} \] Therefore: \[ (4x^4 y^{-4})^3 = 64x^{12}y^{-12} \]
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Simplify the denominator: \[ (x^7 y^3)^2 = (x^7)^2 (y^3)^2 \] Now calculate each part: \[ (x^7)^2 = x^{14}, \quad (y^3)^2 = y^{6} \] Therefore: \[ (x^7 y^3)^2 = x^{14} y^{6} \]
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Combine the results: Now combine the results from the numerator and denominator: \[ \frac{64x^{12}y^{-12}}{x^{14}y^{6}} \]
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Apply the properties of exponents: Using the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\), we can simplify: \[ = 64 \cdot \frac{x^{12}}{x^{14}} \cdot \frac{y^{-12}}{y^{6}} \] Thus: \[ = 64 \cdot x^{12 - 14} \cdot y^{-12 - 6} \] Which simplifies to: \[ = 64 \cdot x^{-2} \cdot y^{-18} \]
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Final expression: Since \(x^{-2} = \frac{1}{x^2}\) and \(y^{-18} = \frac{1}{y^{18}}\), we can write the final expression as: \[ = \frac{64}{x^2 y^{18}} \]
Thus, the equivalent algebraic expression is:
\[ \frac{64}{x^2 y^{18}} \]
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