how would you solve this problem? (3x^2 y)^-3 (3^3 x^-4 y^4)^-5 sorry, I don't know how to do superscript, so the ^ are to represent the exponents.
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Originally Posted by mackenzie25 how would you solve these problems? (3x^2 y)^-3 (3^3 x^-4 y^4)^-5 sorry, I don't know how to do superscript, so the ^ are to represent the exponents. Hi mackenzie25, $\displaystyle \frac{(3x^2y)^{-3}}{(3^3x^{-4}y^4)^{-5}}=\frac{(3^3x^{-4}y^4)^5}{(3x^2y)^3}=\frac{3^{15}x^{-20}y^{20}}{3^3 x^6 y^3}=3^{12}x^{-26}y^{17}=\frac{3^{12}y^{17}}{x^{26}}=\frac{531441 y^{17}}{x^{26}}$
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