Question 291141
root of -8 / x, the problem has a little 3 in front of the root sign. I need to rationalize the denominator. 

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{{{root(3,-8/x)}}}

First we can always take a negative out of a radical if the
index of the radical is odd. The index is 3 (That's the 
little 3 in front of the root sign) and therefore odd. So
let's take the negative sign outside:

{{{-root(3,8/x)}}}

So we need to get an exponent of 3 in the denominator.  There is
an {{{x}}} there, but we need there to be an {{{x^3}}} there, so
we multiply top and bottom by {{{x*x}}}

{{{-root(3,(8*x*x)/(x*x*x))}}}

{{{-root(3,(8*x^2)/(x^3))}}}

Now we use the principle "The cube root of a quotient equals
the quotient of cube roots".

{{{-root(3,8*x^2)/root(3,x^3)}}}  

The denominator is now the cube root of a cube, so
that takes away the radical and the exponent, and just
leaves x in the bottom.

{{{-root(3,8*x^2)/x}}}

Only one thing left to do, and that is to take the 
cube root of 8 which is 2, and put 2 on the outside
of the cube root:

{{{-(2root(3,x^2))/x}}}