Answers: 1
![3ab^2\sqrt[3]{b}](/tpl/images/0078/8114/e756d.png)
![\sqrt[3]{27a^3b^7}](/tpl/images/0078/8114/a44d0.png)
.![\sqrt[3]{27}=(3)](/tpl/images/0078/8114/39d68.png)
since 
.![\sqrt[3]{a^3}=(a)](/tpl/images/0078/8114/d03bb.png)
since 
![\sqrt[3]{b^7}](/tpl/images/0078/8114/5834d.png)
. This problem is a little harder because 
is not a perfect cubes like the others were. But 
so rewrite 
or 
.![\sqrt[3]{b^6 \cdot b}=\sqrt[3]{b^6}\cdot \sqrt[3]{b}=b^2 \cdot \sqrt[3]{b}](/tpl/images/0078/8114/1efb8.png)
. The 
came from this 
.
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