How do you simplify (a^3 - b^3)/(3a^2 + 9ab + 6b^2)*(a^2 + 2ab + b^2)/(a^2 - b^2)?

Jul 19, 2015

Factor and cancel matching factors to find:

$\frac{{a}^{3} - {b}^{3}}{3 {a}^{2} + 9 a b + 6 {b}^{2}} \cdot \frac{{a}^{2} + 2 a b + {b}^{2}}{{a}^{2} - {b}^{2}} = \frac{{a}^{2} + a b + {b}^{2}}{3 \left(a + 2 b\right)}$

Explanation:

Use difference of cubes identity:

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

Use difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$\frac{{a}^{3} - {b}^{3}}{3 {a}^{2} + 9 a b + 6 {b}^{2}} \cdot \frac{{a}^{2} + 2 a b + {b}^{2}}{{a}^{2} - {b}^{2}}$

$= \frac{\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) {\left(a + b\right)}^{2}}{3 \left({a}^{2} + 3 a b + 2 {b}^{2}\right) \left(a - b\right) \left(a + b\right)}$

$= \frac{\left({a}^{2} + a b + {b}^{2}\right) \left(a + b\right)}{3 \left(a + b\right) \left(a + 2 b\right)}$

$= \frac{{a}^{2} + a b + {b}^{2}}{3 \left(a + 2 b\right)}$