How do you simplify $\frac{x^{3} + 27}{9 x + 27} / \frac{3 x^{2} - 9 x + 27}{4 x}$ $(x^3+27)/(9x+27) / (3x^2-9x+27)/(4x)$ ?

Question

1951 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-23T04:51:57

$\frac{4 x}{27}$ $(4x)/27$

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

$= \frac{x^{3} + 27}{9 x + 27} (\frac{4 x}{3 x^{2} - 9 x + 27})$ $=(x^3+27)/(9x+27)((4x)/(3x^2-9x+27))$

Factor each term.

The top left is a sum of cubes.

$= \frac{(x + 3) (x^{2} - 3 x + 9)}{9 x + 27} (\frac{4 x}{3 x^{2} - 9 x + 27})$ $=((x+3)(x^2-3x+9))/(9x+27)((4x)/(3x^2-9x+27))$

Factor a $9$ $9$ from the bottom left.

$= \frac{(x + 3) (x^{2} - 3 x + 9)}{9 (x + 3)} (\frac{4 x}{3 x^{2} - 9 x + 27})$ $=((x+3)(x^2-3x+9))/(9(x+3))((4x)/(3x^2-9x+27))$

Cancel the $(x + 3)$ $(x+3)$ terms.

$= \frac{x^{2} - 3 x + 9}{9} (\frac{4 x}{3 x^{2} - 9 x + 27})$ $=(x^2-3x+9)/(9)((4x)/(3x^2-9x+27))$

Factor a $3$ $3$ from the bottom right.

$= \frac{x^{2} - 3 x + 9}{9} (\frac{4 x}{3 (x^{2} - 3 x + 9)})$ $=(x^2-3x+9)/(9)((4x)/(3(x^2-3x+9)))$

Notice that the $x^{2} - 3 x + 9$ $x^2-3x+9$ terms will cancel.

$= \frac{1}{9} (\frac{4 x}{3}) = \frac{4 x}{27}$ $=1/9((4x)/3)=(4x)/27$

How do you simplify x3+279x+27/3x2−9x+274x (x^3+27)/(9x+27) / (3x^2-9x+27)/(4x)?