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.

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

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