Question 189564
One thing to note: {{{root(3,4x)=root(3,2x*2x)=root(3,2x)root(3,2x)}}}



{{{2/root(3,2x)}}} Start with the given expression.



To rationalize the denominator, we need the denominator to be the cube root of something cubed (so the cube root would cancel out). So we need 3 instances of {{{2x}}} to get {{{(2x)^3}}}. This is why we multiply both the numerator and denominator by {{{root(3,2x)root(3,2x)}}} (add these to the original root to get 3 roots)




{{{(2*root(3,2x)root(3,2x))/(root(3,2x)root(3,2x)root(3,2x))}}} Multiply both the numerator and denominator by {{{root(3,2x)root(3,2x)}}}



{{{(2*root(3,2x*2x))/(root(3,2x*2x*2x))}}} Combine the roots.



{{{(2*root(3,(2x)^2))/(root(3,(2x)^3))}}} Condense the terms.



{{{(2*root(3,(2x)^2))/(2x)}}} Take the cube root of the cube in the denominator to cancel out the cube root. The denominator is now rationalized.