# How do find the quotient of (y^3 - 125)/(y - 5)?

Oct 30, 2015

$\left({y}^{2} + 5 y + 25\right)$

#### Explanation:

You need to factor in order to get rid of the denominator. In order to do that you need to use the difference of cubes. That is

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab + b2)

We are going to use the second one

Also $a = y , b = 5$

Turn ${y}^{3} - 125$ to ${y}^{3} - {5}^{3}$

Use the formula

${y}^{3} - {5}^{3} = \left(y - 5\right) \left({y}^{2} + 5 y + 25\right)$

Plug that into the numerator

$\frac{\left(y - 5\right) \left({y}^{2} + 5 y + 25\right)}{\left(y - 5\right)}$

Simplify

$\left({y}^{2} + 5 y + 25\right)$