How do you rationalize the denominator and simplify #root3( (1 / (2x^2)))#?

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Answer 1 · 2017-08-20T19:16:16

You #ul("force")# it into something you can take the cube root of.

#=root(3)(4x)/(2x)#

Multiply by 1 and you do not change the intrinsic value. However, 1 comes in many forms

#color(green)(root(3)(1/(2x^2)color(red)(xx1)))#

#2xx2^2=2^3#

#x^2xx x^2xx x^2 =x^2xx(x^2)^2 = (x^2)^3#

#color(green)(root(3)(1/(2x^2)color(red)(xx(2^2(x^2)^2)/(2^2(x^2)^2))))" "->" "root(3)((4x^4)/(2^3(x^2)^3))#

#color(white)("vvvvvvvvvvvvvvv")->" "root(3)(4x^4)/(2x^2)#
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Simplified further

#x^4->x^3xx x" "# so we can change the #4x^4# such that we have:

#(xroot(3)4x)/(2x^2) = x/x xxroot(3)(4x)/(2x)#

#=root(3)(4x)/(2x)#