# How do you simplify (16x^3 - 2y^3)/(16x^3 + 8x^2 y + 4xy^2)?

Mar 2, 2018

$= \frac{2 x - y}{2 x}$

#### Explanation:

You can only cancel if you have factors.

Factorise top and bottom first by taking out the common factor.

(16x^3-2y^3)/(16x^3 +8x^2y+4xy^2) = (2(8x^3-y^3))/(4x(4x^2+2xy+y^2)

THe numerator can be factored further as the difference of cubes:

$\frac{2 \textcolor{b l u e}{\left(8 {x}^{3} - {y}^{3}\right)}}{4 x \left(4 {x}^{2} + 2 x y + {y}^{2}\right)}$

 = (2color(blue)((2x-y)(4x^2+2xy+y^2)))/(4x(4x^2+2xy+y^2)

Now cancel the factors which are the same.

 = (cancel2(2x-y)cancel((4x^2+2xy+y^2)))/(cancel4^2xcancel((4x^2+2xy+y^2))

$= \frac{2 x - y}{2 x}$

Recall: Difference of cubes:

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