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

Question

Isaiah Anthony · Accepted Answer

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#

Julia Adams · Answer

undefined To divide (x^3+27)/(9x+27) by (3x^2-9x+27)/(4x), we can follow these steps:

1. Factorize the numerator and denominator of both fractions, if possible.
   (x^3+27) can be factored as (x+3)(x^2-3x+9).
   (9x+27) can be factored as 9(x+3).
   (3x^2-9x+27) can be factored as 3(x^2-3x+9).
   (4x) can be factored as 4(x).

2. Rewrite the division as multiplication by the reciprocal of the second fraction.
   (x^3+27)/(9x+27) * (4x)/(3x^2-9x+27)

3. Cancel out any common factors between the numerators and denominators.
   (x+3) cancels out in the numerator and denominator.

4. Multiply the remaining factors together.
   (x^2-3x+9) * (4x) / (9 * 3(x^2-3x+9))

5. Simplify the expression.
   (4x(x^2-3x+9)) / (27(x^2-3x+9))

6. Cancel out any common factors between the numerator and denominator.
   (x^2-3x+9) cancels out.

7. The final result is:
   4x / 27