# Is x^12-y^12 difference of two squares or difference of two cubes?

##### 1 Answer
Apr 17, 2015

It could be both, actually.

You can use the properties of exponential powers to write those terms both as a difference of squares, and as a difference of cubes.

Since ${\left({a}^{x}\right)}^{y} = {a}^{x y}$, you can say that

${x}^{12} = {x}^{6 \cdot \textcolor{red}{2}} = {\left({x}^{6}\right)}^{\textcolor{red}{2}}$

and

y^(12) = (y^(6))^(color(red)(2)

This means that you get

${x}^{12} - {y}^{12} = {\left({x}^{6}\right)}^{2} - {\left({y}^{6}\right)}^{2} = \left({x}^{6} - {y}^{6}\right) \left({x}^{6} + {y}^{6}\right)$

Likewise,

${x}^{12} = {x}^{4 \cdot \textcolor{red}{3}} = {\left({x}^{4}\right)}^{\textcolor{red}{3}}$ and ${y}^{12} = {\left({y}^{4}\right)}^{\textcolor{red}{3}}$

So you can write

${x}^{12} - {y}^{12} = {\left({x}^{4}\right)}^{3} - {\left({y}^{4}\right)}^{3} = \left({x}^{4} - {y}^{4}\right) \left[{\left({x}^{4}\right)}^{2} + {x}^{4} {y}^{4} + {\left({y}^{4}\right)}^{2}\right]$

${x}^{12} - {y}^{12} = \left({x}^{4} - {y}^{4}\right) \left[{x}^{8} + {x}^{4} {y}^{4} + {y}^{8}\right]$

As you can see, you can simplify these expressions further. Here's how you would factor this expression completely

${x}^{12} - {y}^{12} = {\underbrace{\left({x}^{6} - {y}^{6}\right)}}_{\textcolor{g r e e n}{\text{difference of two squares")) * underbrace((x^6 + y^6))_(color(blue)("sum of two cubes}}} =$

$= {\underbrace{\left({x}^{3} - {y}^{3}\right)}}_{\textcolor{g r e e n}{\text{difference of two cubes")) * underbrace((x^3 + y^3))_(color(blue)("sum of two cubes}}} \cdot \left({x}^{2} + {y}^{2}\right) \left({x}^{4} + {x}^{2} \cdot {y}^{2} + {y}^{4}\right) =$

$= \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right) \cdot \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right) \cdot \left({x}^{2} + {y}^{2}\right) \left({x}^{4} + {x}^{2} \cdot {y}^{2} + {y}^{4}\right)$

${x}^{12} - {y}^{12} = \left(x + y\right) \left(x - y\right) \left({x}^{2} + {y}^{2}\right) \left({x}^{2} - x y + {y}^{2}\right) \left({x}^{2} + x y + {y}^{2}\right) \left({x}^{4} + {x}^{2} {y}^{2} + {y}^{2}\right)$