# How do you factor 27k^3+64d^3?

Aug 29, 2016

$27 {k}^{3} + 64 {d}^{3} = \left(3 k + 4 d\right) \left(9 {k}^{2} - 12 k d + 16 {d}^{2}\right)$

#### Explanation:

The sum of cubes identity can be written:

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

We can use this with $a = 3 k$ and $b = 4 d$ to find:

$27 {k}^{3} + 64 {d}^{3}$

$= {\left(3 k\right)}^{3} + {\left(4 d\right)}^{3}$

$= \left(3 k + 4 d\right) \left({\left(3 k\right)}^{2} - \left(3 k\right) \left(4 d\right) + {\left(4 d\right)}^{2}\right)$

$= \left(3 k + 4 d\right) \left(9 {k}^{2} - 12 k d + 16 {d}^{2}\right)$

That is as far as we can go with Real coefficients. If we allow Complex coefficients then this can be factored further as:

$= \left(3 k + 4 d\right) \left(3 k + 4 \omega d\right) \left(3 k + 4 {\omega}^{2} d\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$

Aug 29, 2016

=$\left(3 k + 4 d\right) \left(9 {k}^{2} - 12 \mathrm{dk} + 16 {d}^{2}\right)$

#### Explanation:

This difficulty is recognising that the values are all cubes.

This expression is the sum of cubes.

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

$27 {k}^{3} + 64 {d}^{3} = \left(3 k + 4 d\right) \left(9 {k}^{2} - 12 \mathrm{dk} + 16 {d}^{2}\right)$