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how do we solve ((81x^-4)/(16y^4))^-(3/4) ?

Guest Sep 30, 2014

Best Answer 

 #1
avatar+26969 
+10

There is nothing to solve, but we can simplify it:

 

$$(\frac{81x^{-4}}{16y^4})^{-\frac{3}{4}}\\\\
(\frac{81}{16x^4y^4})^{-\frac{3}{4}}\\\\
(\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}x^{4\frac{1}{4}}y^{4\frac{1}{4}}})^{-3}\\\\
(\frac{3}{2xy})^{-3}\\\\
\frac{2^3x^3y^3}{3^3}\\\\
\frac{8x^3y^3}{27}$$

Alan  Sep 30, 2014
 #1
avatar+26969 
+10
Best Answer

There is nothing to solve, but we can simplify it:

 

$$(\frac{81x^{-4}}{16y^4})^{-\frac{3}{4}}\\\\
(\frac{81}{16x^4y^4})^{-\frac{3}{4}}\\\\
(\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}x^{4\frac{1}{4}}y^{4\frac{1}{4}}})^{-3}\\\\
(\frac{3}{2xy})^{-3}\\\\
\frac{2^3x^3y^3}{3^3}\\\\
\frac{8x^3y^3}{27}$$

Alan  Sep 30, 2014

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