# How i can factor 8*(a^(6n))+27*(b^(3m)) ?

Sep 9, 2015

$\left(2 {a}^{2 n} + 3 {b}^{m}\right) \cdot \left(4 {a}^{4 n} - 6 {a}^{2 n} {b}^{m} + 9 {b}^{2 m}\right)$

#### Explanation:

Here's what I'd try.

Notice that you can write

$8 = 2 \text{^3" }$ and ${\text{ "27 = 3}}^{3}$

which means that you can rewrite the original expression as

${2}^{3} \cdot {a}^{6 n} + {3}^{3} \cdot {b}^{3 m}$

You can do the same for

${a}^{6 n} = {\left({a}^{2 n}\right)}^{3} \text{ }$ and $\text{ } {b}^{3 m} = {\left({b}^{m}\right)}^{3}$

This will give you

${2}^{3} \cdot {\left({a}^{2 n}\right)}^{3} + {3}^{3} \cdot {\left({b}^{m}\right)}^{3} = {\left(2 \cdot {a}^{2 n}\right)}^{3} + {\left(3 \cdot {b}^{m}\right)}^{3}$

Now you can use the sum of cubes factoring formula

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

to get

$\left(2 {a}^{2 n} + 3 {b}^{m}\right) \cdot \left[{\left(2 {a}^{2 n}\right)}^{2} - 2 {a}^{2 n} \cdot 3 {b}^{m} + {\left(3 {b}^{m}\right)}^{2}\right]$

$\left(2 {a}^{2 n} + 3 {b}^{m}\right) \cdot \left(4 {a}^{4 n} - 6 {a}^{2 n} {b}^{m} + 9 {b}^{2 m}\right)$