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

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Answer 1 · 2015-05-26T23:23:46

There seems to be a couple of #(x+3)#'s going on here...

#x^3+27 = x^3+3^3#

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

...using the sum of cubes identity

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

#9x + 27 = 9(x+3)#

So:

#(x^3 + 27)/(9x+27)#

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

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

Then the numerator of the second factor is

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

So we have:

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

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

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

with the proviso that #x != -3# and #x != 0#