# How do you factor (3y + 1)^3 + 8y^3?

Aug 20, 2017

${\left(3 y + 1\right)}^{3} + 8 {y}^{3} = \left(5 y + 1\right) \left(7 {y}^{2} + 4 y + 1\right)$

$\textcolor{w h i t e}{{\left(3 y + 1\right)}^{3} + 8 {y}^{3}} = \frac{1}{7} \left(5 y + 1\right) \left(7 y + 2 - \sqrt{3} i\right) \left(7 y + 2 + \sqrt{3} i\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Since both ${\left(3 y + 1\right)}^{3}$ and $8 {y}^{3} = {\left(2 y\right)}^{3}$ are perfect cubes, we can use the sum of cubes identity with $a = \left(3 y + 1\right)$ and $b = 2 y$ as follows:

${\left(3 y + 1\right)}^{3} + 8 {y}^{3} = {\left(3 y + 1\right)}^{3} + {\left(2 y\right)}^{3}$

$\textcolor{w h i t e}{{\left(3 y + 1\right)}^{3} + 8 {y}^{3}} = \left(\left(3 y + 1\right) + 2 y\right) \left({\left(3 y + 1\right)}^{2} - \left(3 y + 1\right) \left(2 y\right) + {\left(2 y\right)}^{2}\right)$

$\textcolor{w h i t e}{{\left(3 y + 1\right)}^{3} + 8 {y}^{3}} = \left(5 y + 1\right) \left(\left(9 {y}^{2} + 6 y + 1\right) - \left(6 {y}^{2} + 2 y\right) + 4 {y}^{2}\right)$

$\textcolor{w h i t e}{{\left(3 y + 1\right)}^{3} + 8 {y}^{3}} = \left(5 y + 1\right) \left(7 {y}^{2} + 4 y + 1\right)$

The remaining quadratic factor has no linear factors with real coefficients.

$\textcolor{w h i t e}{}$
Complex factors

If we do want to factor $7 {y}^{2} + 4 x + 1$ then we can do it by completing the square and using the difference of squares identity, with non-real complex coefficients.

The difference of squares identity can be written:

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

We can use this with $a = \left(7 y + 2\right)$ and $b = \sqrt{3} i$ as follows:

$7 \left(7 {y}^{2} + 4 y + 1\right) = 49 {y}^{2} + 28 y + 7$

$\textcolor{w h i t e}{7 \left(7 {y}^{2} + 4 y + 1\right)} = {\left(7 y\right)}^{2} + 2 \left(7 y\right) \left(2\right) + {2}^{2} + 3$

$\textcolor{w h i t e}{7 \left(7 {y}^{2} + 4 y + 1\right)} = {\left(7 y + 2\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{7 \left(7 {y}^{2} + 4 y + 1\right)} = \left(\left(7 y + 2\right) - \sqrt{3} i\right) \left(\left(7 y + 2\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{7 \left(7 {y}^{2} + 4 y + 1\right)} = \left(7 y + 2 - \sqrt{3} i\right) \left(7 y + 2 + \sqrt{3} i\right)$

Dividing both ends by $7$ we get:

$7 {y}^{2} + 4 y + 1 = \frac{1}{7} \left(7 y + 2 - \sqrt{3} i\right) \left(7 y + 2 + \sqrt{3} i\right)$