+0

# Simplfying radicals

+1
110
2
+701

http://prntscr.com/l1s9ud

Oct 3, 2018

### 2+0 Answers

#1
+3836
+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 $$\boxed{2\sqrt[3]{2}\sqrt[3]{x^5y^{13}}}$$.

Oct 3, 2018
#2
+16321
+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 $$\sqrt[3]{2x^2 y}$$

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Oct 3, 2018
edited by ElectricPavlov  Oct 3, 2018

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