How do you simplify (2a^2+ab-b^2)/(a^3+b^3) ÷ (2a^2b^2-ab^3)/(2a^2-2ab+2b^2)?

Mar 3, 2017

$= \frac{2}{a {b}^{2}}$

Explanation:

The first step in algebraic fractions is to factorise wherever possible.

color(red)((2a^2+ab-b^2))/color(green)((a^3+b^3)) ÷ color(blue)((2a^2b^2-ab^3))/color(purple)((2a^2-2ab+2b^2))

$\textcolor{red}{\text{quadratic trinomial")/color(green)("sum of cubes") ÷ color(blue)("common factor")/color(purple)("common factor}}$

=color(red)((2a-b)(a+b))/color(green)((a+b)(a^2-ab+b^2)) ÷ color(blue)(ab^2(2a -b))/color(purple)(2(a^2-ab+b^2))

To divide: multiply by the reciprocal

$= \frac{\textcolor{red}{\left(2 a - b\right) \left(a + b\right)}}{\textcolor{g r e e n}{\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)}} \times \frac{\textcolor{p u r p \le}{2 \left({a}^{2} - a b + {b}^{2}\right)}}{\textcolor{b l u e}{a {b}^{2} \left(2 a - b\right)}}$

Cancel like factors

$= \frac{\cancel{\left(2 a - b\right)} \cancel{\left(a + b\right)}}{\cancel{\left(a + b\right)} \cancel{\left({a}^{2} - a b + {b}^{2}\right)}} \times \frac{\textcolor{p u r p \le}{2 \cancel{\left({a}^{2} - a b + {b}^{2}\right)}}}{a {b}^{2} \cancel{\left(2 a - b\right)}}$

$= \frac{2}{a {b}^{2}}$