# How do you simplify (-91a^4b+14ab )/ (-7ab)?

Apr 19, 2016

$\frac{- 91 {a}^{4} b + 14 a b}{- 7 a b}$

$= 13 {a}^{3} - 2$

$= \left(\sqrt[3]{13} a - \sqrt[3]{2}\right) \left(\sqrt[3]{169} {a}^{2} + \sqrt[3]{26} a + \sqrt[3]{4}\right)$

with exclusions $a \ne 0$ and $b \ne 0$

#### Explanation:

Both of the terms $- 91 {a}^{4} b$ and $14 a b$ are divisible by $- 7 a b$, so we find:

$\frac{- 91 {a}^{4} b + 14 a b}{- 7 a b} = 13 {a}^{3} - 2$

with exclusions $a \ne 0$ and $b \ne 0$

Note that the exclusions are required because if $a = 0$ or $b = 0$ then the original expression is of the form $\frac{0}{0}$ which is undefined.

I guess that counts as 'simplified', but it can be factored too, using cube roots...

For any Real numbers $x$ and $y$ note that:

$\sqrt[3]{x} \sqrt[3]{y} = \sqrt[3]{x y}$

We will use this below.

The difference of cubes identity can be written:

${A}^{3} - {B}^{3} = \left(A - B\right) \left({A}^{2} + A B + {B}^{2}\right)$

We can factor $13 {a}^{3} - 2$ using this identity with $A = \sqrt[3]{13} a$ and $b = \sqrt[3]{2}$ as follows:

$13 {a}^{3} - 2$

$= {\left(\sqrt[3]{13} a\right)}^{3} - {\left(\sqrt[3]{2}\right)}^{3}$

$= \left(\sqrt[3]{13} a - \sqrt[3]{2}\right) \left({\left(\sqrt[3]{13} a\right)}^{2} + \left(\sqrt[3]{13} a\right) \left(\sqrt[3]{2}\right) + {\left(\sqrt[3]{2}\right)}^{2}\right)$

$= \left(\sqrt[3]{13} a - \sqrt[3]{2}\right) \left(\sqrt[3]{169} {a}^{2} + \sqrt[3]{26} a + \sqrt[3]{4}\right)$