# How do you factor 64 - c^12?

Dec 15, 2015

Use some special identities to find:

$64 - {c}^{12} = \left(\sqrt{2} - c\right) \left(\sqrt{2} + c\right) \left(2 + {c}^{2}\right) \left(2 + \sqrt{2} c + {c}^{2}\right) \left(2 - \sqrt{2} c + {c}^{2}\right) \left(2 + \sqrt{6} c + {c}^{2}\right) \left(2 - \sqrt{6} c + {c}^{2}\right)$

#### Explanation:

The difference of cubes identity can be written:

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

The difference of squares identity can be written:

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

Note also that:

$\left({a}^{2} + a b + {b}^{2}\right) \left({a}^{2} - a b + {b}^{2}\right) = {a}^{4} + {a}^{2} {b}^{2} + {b}^{4}$

The final identity we will use is:

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

These last two identities were found by looking at:

$\left({a}^{2} + k a b + {b}^{2}\right) \left({a}^{2} - k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

and picking $k = 1$ or $k = \sqrt{3}$ to get the required result.

Note that $64 = {4}^{3}$ and ${c}^{12} = {\left({c}^{4}\right)}^{3}$ are both perfect cubes. So we can use the difference of cubes identity to start to factorise this.

Letting $a = 4$ and $b = {c}^{4}$ we find:

$64 - {c}^{12}$

$= {4}^{3} - {\left({c}^{4}\right)}^{3}$

$= \left(4 - {c}^{4}\right) \left({4}^{2} + 4 {c}^{4} + {\left({c}^{4}\right)}^{2}\right)$

$= \left(4 - {c}^{4}\right) \left(16 + 4 {c}^{4} + {c}^{8}\right)$

Next both $4 = {2}^{2}$ and ${c}^{4} = {\left({c}^{2}\right)}^{2}$ are perfect squares so we can factor some more:

$= \left({2}^{2} - {\left({c}^{2}\right)}^{2}\right) \left(16 + 4 {c}^{4} + {c}^{8}\right)$

$= \left(2 - {c}^{2}\right) \left(2 + {c}^{2}\right) \left(16 + 4 {c}^{4} + {c}^{8}\right)$

Next use the difference of squares identity to factor the first quadratic:

$= \left({\sqrt{2}}^{2} - {c}^{2}\right) \left(2 + {c}^{2}\right) \left(16 + 4 {c}^{4} + {c}^{8}\right)$

$= \left(\sqrt{2} - c\right) \left(\sqrt{2} + c\right) \left(2 + {c}^{2}\right) \left(16 + 4 {c}^{4} + {c}^{8}\right)$

Next use our special ${a}^{4} + {a}^{2} {b}^{2} + {b}^{4}$ identity with $a = 2$ and $b = {c}^{2}$ to factor some more:

$= \left(\sqrt{2} - c\right) \left(\sqrt{2} + c\right) \left(2 + {c}^{2}\right) \left(4 + 2 {c}^{2} + {c}^{4}\right) \left(4 - 2 {c}^{2} + {c}^{4}\right)$

Then use it again with $a = \sqrt{2}$ and $b = c$ to factor some more:

$= \left(\sqrt{2} - c\right) \left(\sqrt{2} + c\right) \left(2 + {c}^{2}\right) \left(2 + \sqrt{2} c + {c}^{2}\right) \left(2 - \sqrt{2} c + {c}^{2}\right) \left(4 - 2 {c}^{2} + {c}^{4}\right)$

Then use our final identity with $a = \sqrt{2}$ and $b = c$ to factor the last quartic:

$= \left(\sqrt{2} - c\right) \left(\sqrt{2} + c\right) \left(2 + {c}^{2}\right) \left(2 + \sqrt{2} c + {c}^{2}\right) \left(2 - \sqrt{2} c + {c}^{2}\right) \left(2 + \sqrt{6} c + {c}^{2}\right) \left(2 - \sqrt{6} c + {c}^{2}\right)$

That's as far as we can get with Real coefficients.