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

May 26, 2015

There seems to be a couple of $\left(x + 3\right)$'s going on here...

${x}^{3} + 27 = {x}^{3} + {3}^{3}$

$= \left(x + 3\right) \left({x}^{2} - 3 x + {3}^{2}\right) = \left(x + 3\right) \left({x}^{2} - 3 x + 9\right)$

...using the sum of cubes identity

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$9 x + 27 = 9 \left(x + 3\right)$

So:

$\frac{{x}^{3} + 27}{9 x + 27}$

$= \frac{\cancel{x + 3} \left({x}^{2} - 3 x + 9\right)}{9 \left(\cancel{x + 3}\right)}$

$= \frac{{x}^{2} - 3 x + 9}{9}$

Then the numerator of the second factor is

$3 {x}^{2} - 9 x + 27 = 3 \left({x}^{2} - 3 x + 9\right)$

So we have:

$\frac{{x}^{3} + 27}{9 x + 27} \div \frac{3 {x}^{2} - 9 x + 27}{4 x}$

$= \frac{\cancel{{x}^{2} - 3 x + 9}}{9} \div \frac{3 \cdot \cancel{{x}^{2} - 3 x + 9}}{4 x}$

$= \frac{4 x}{9 \cdot 3} = \frac{4 x}{27}$

with the proviso that $x \ne - 3$ and $x \ne 0$