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http://prntscr.com/l1s9ud

critical  Oct 3, 2018
 #1
avatar+3171 
+2

You can break the original value into \(\sqrt[3]{16}\sqrt[3]{x^5y^{13}}.\) We know that \(\sqrt[3]{16}=2\sqrt[3]{2}\), because of the power,  \(2^4.\) Thus, the answer is \(\boxed{2\sqrt[3]{2}\sqrt[3]{x^5y^{13}}}\).

tertre  Oct 3, 2018
 #2
avatar+13027 
+2

....further   y^13 =  y^3 y^3 y^3 y^3 y     so you can take all of those y^3 's out

   and   x^5 = x^3 x^2       so you can take the x^3 out

 

2 x y^4 \(\sqrt[3]{2x^2 y}\)

ElectricPavlov  Oct 3, 2018
edited by ElectricPavlov  Oct 3, 2018

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