Question 769516
Simplify {{{root(3,-16xy^6)}}}

A.  -2y^2{{{root(3,2x)}}}
B.  2y^2{{{root(3,-x)}}}
C.  2y^2{{{root(3,-2x)}}}
D.  -4y^2{{{root(3,x)}}}

I thought it was C.  {{{root(3,16xy^6)}}} = {{{root(3,-2x*8y^6)}}}={{{root(3,-2x)}}}{{{root(3,8y^6)}}}=2y^2{{{root(3,-2x)}}}

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Since it's an odd root, 
= {{{-2y^2root(3,2x)}}}
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Both answers are correct, but there's less under the radical when the -1 is removed.