# How do you factor 8a^3 + 27b^3 + 2a + 3b?

$8 {a}^{3} + 27 {b}^{3} + 2 a + 3 b = \left(2 a + 3 b\right) \left(4 {a}^{2} - 6 a b + 9 {b}^{2} + 1\right)$

#### Explanation:

From the given expression
$8 {a}^{3} + 27 {b}^{3} + 2 a + 3 b$

Factoring by grouping

$8 {a}^{3} + 27 {b}^{3} + 2 a + 3 b$

$\left(8 {a}^{3} + 27 {b}^{3}\right) + \left(2 a + 3 b\right)$

The first two terms can be factored by sum of two cubes formula

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

so that $8 {a}^{3} + 27 {b}^{3} = \left(2 a + 3 b\right) \left(4 {a}^{2} - 6 a b + 9 {b}^{2}\right)$

Let us continue

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

factor out the common term (2a+3b)

$\left(2 a + 3 b\right) \left(4 {a}^{2} - 6 a b + 9 {b}^{2} + 1\right)$

therefore

$8 {a}^{3} + 27 {b}^{3} + 2 a + 3 b = \left(2 a + 3 b\right) \left(4 {a}^{2} - 6 a b + 9 {b}^{2} + 1\right)$

God bless...I hope the explanation is useful.