How do you simplify # (x^3+27)/(9x+27) / (3x^2-9x+27)/(4x)#?

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Answer 1 · 2016-01-23T04:51:57

#(4x)/27#

Recall that division is the same as multiplying by the reciprocal. Thus, we can flip the right term and multiply instead of divide.

#=(x^3+27)/(9x+27)((4x)/(3x^2-9x+27))#

Factor each term.

The top left is a sum of cubes.

#=((x+3)(x^2-3x+9))/(9x+27)((4x)/(3x^2-9x+27))#

Factor a #9# from the bottom left.

#=((x+3)(x^2-3x+9))/(9(x+3))((4x)/(3x^2-9x+27))#

Cancel the #(x+3)# terms.

#=(x^2-3x+9)/(9)((4x)/(3x^2-9x+27))#

Factor a #3# from the bottom right.

#=(x^2-3x+9)/(9)((4x)/(3(x^2-3x+9)))#

Notice that the #x^2-3x+9# terms will cancel.

#=1/9((4x)/3)=(4x)/27#