# How do you factor 1331s^12 - 1728t^6?

May 15, 2016

$1331 {s}^{12} - 1728 {t}^{6}$

$= \left(\sqrt{11} {s}^{2} - 2 \sqrt{3} t\right) \left(11 {s}^{4} + 2 \sqrt{33} {s}^{2} t + 12 {t}^{2}\right) \left(\sqrt{11} {s}^{2} + 2 \sqrt{3} t\right) \left(11 {s}^{4} - 2 \sqrt{33} {s}^{2} t + 12 {t}^{2}\right)$

#### Explanation:

We will use the following identities:

• Difference of squares: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

• Difference of cubes: ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

• Sum of cubes: ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

First let us find out how to factorise ${a}^{6} - {b}^{6}$

We can treat this as a difference of squares, then difference and sum of cubes as follows:

${a}^{6} - {b}^{6}$

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

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

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

So we have our very own difference of sixth powers identity:

${a}^{6} - {b}^{6} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Note that:

• $1331 = {11}^{3} = {\left(\sqrt{11}\right)}^{6}$

• $1728 = {12}^{3} = {\left({2}^{2} \cdot 3\right)}^{3} = {\left(2 \sqrt{3}\right)}^{6}$

Let's put our difference of sixth powers to work with $a = \sqrt{11} {s}^{2}$ and $b = 2 \sqrt{3} t$

${a}^{6} - {b}^{6}$

$= {\left(\sqrt{11} {s}^{2}\right)}^{6} - {\left(2 \sqrt{3} t\right)}^{6}$

$= {11}^{3} {s}^{12} - {12}^{3} {t}^{6}$

$= 1331 {s}^{12} - 1728 {t}^{6}$

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

$= \left(\sqrt{11} {s}^{2} - 2 \sqrt{3} t\right) \left(11 {s}^{4} + 2 \sqrt{33} {s}^{2} t + 12 {t}^{2}\right) \left(\sqrt{11} {s}^{2} + 2 \sqrt{3} t\right) \left(11 {s}^{4} - 2 \sqrt{33} {s}^{2} t + 12 {t}^{2}\right)$