# How do you find the quotient of (c^3-27)div(c-3)?

Mar 28, 2017

${c}^{2} + 3 c + 9$

#### Explanation:

we use a long division to solve it.

${c}^{3} - 27$ $- - \to {c}^{2} \cdot \left(c - 3\right)$
$- \left({c}^{3} - 3 {c}^{2}\right)$
.................................
$3 {c}^{2} - 27$ $- - \to 3 c \cdot \left(c - 3\right)$
$- \left(3 {c}^{2} - 9 c\right)$
.................................
$9 c - 27$ $- - \to 9 \cdot \left(c - 3\right)$
$- \left(9 c - 27\right)$
...............................
$0$

the answer is ${c}^{2} + 3 c + 9$

Mar 28, 2017

${c}^{2} + 3 c + 9$

#### Explanation:

The numerator is a $\textcolor{b l u e}{\text{difference of cubes}}$ and in general is factorised as shown.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

${c}^{3} - 27 = {\left(c\right)}^{3} - {\left(3\right)}^{3} \Rightarrow a = c \text{ and } b = 3$

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

rArr(c^3-27)/(c-3)=(cancel((c-3)^1)(c^2+3c+9))/(cancel((c-3)^1)

$= {c}^{2} + 3 c + 9 \leftarrow \textcolor{red}{\text{ quotient}}$