# How do you divide (v^3+27)/(v+3)?

Aug 5, 2015

$\frac{{v}^{3} + 27}{v + 3} = {v}^{2} - 3 v + 9$

#### Explanation:

Assume $v + 3$ is a factor for ${v}^{3} + 27$ and from this infer the remaining factor. This gives:
${v}^{3} + 27 = \left(v + 3\right) \left({v}^{2} - 3 v + 9\right)$

Therefore:
$\frac{{v}^{3} + 27}{v + 3} = {v}^{2} - 3 v + 9$

Aug 6, 2015

${v}^{3} + 27$ is of the form:

${a}^{3} \pm {b}^{3}$

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

or:

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

$a = v$
$b = 3$

So you get, with the first one:

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

$\to \frac{\left(v + 3\right) \left({v}^{2} - 3 v + 9\right)}{v + 3} = \textcolor{b l u e}{{v}^{2} - 3 v + 9}$

No guessing necessary.