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

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

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

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

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

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