Question 201810
note: you can say "root(3,x)" to mean {{{root(3,x)}}}. Or you can simply say "the cube root of 2" to mean {{{root(3,2)}}}




{{{root(3,x)*(root(3,3x^2)-root(3,81x^2))}}} Start with the given expression.



{{{root(3,x)*root(3,3x^2)-root(3,x)*root(3,81x^2)}}} Distribute



{{{root(3,x*3x^2)-root(3,x*81x^2)}}} Combine the roots using the identity {{{root(n,x)*root(n,y)=root(n,x*y)}}} 



{{{root(3,3x^3)-root(3,81x^3)}}} Multiply



{{{root(3,3x^3)-root(3,27*3x^3)}}} Factor 81 into 27*3. Note: one factor must be a perfect cube.



{{{root(3,3)*root(3,x^3)-root(3,27)*root(3,3)*root(3,x^3)}}} Break up the roots.



{{{root(3,3)*root(3,x^3)-3*root(3,3)*root(3,x^3)}}} Evaluate the cube root of 27 to get 3.



{{{root(3,3)*x-3*root(3,3)*x}}} Evaluate the cube root of {{{x^3}}} to get "x".



{{{x*root(3,3)-3x*root(3,3)}}} Rearrange the terms and multiply



{{{(x-3x)*root(3,3)}}} Factor out the GCF {{{root(3,3)}}}



{{{-2x*root(3,3)}}} Combine like terms.




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Answer:



So {{{root(3,x)*(root(3,3x^2)-root(3,81x^2))=-2x*root(3,3)}}}