# How can I simplify this?

$$(x^{2^{n-1}} + y^{2^{n-1}}) (x^{2^{n-1}} - y^{2^{n-1}})$$

I derived it like, $[(x^{2^{n-1}})^2 - (y^{2^{n-1}})^2]$ , as $(a+b) (a-b) = a^2 - b^2$

Is there any way to simplify it further?

-
As a first step to simplify use LaTeX to write mathematics in this site. You can find direction in the FAQ section. As it is now it's very hard to understand it... –  DonAntonio Jan 28 '13 at 12:05
oh, sure. I was wondering the same how I could do that. –  Harish Raj Jan 28 '13 at 12:06
Your formulas don't render appropriately in LaTeX because x^{2}^{n-1} is ambiguous. You should write it with an extra set of braces as x^{2^{n-1}} (rendering as $x^{{2}^{n-1}}$) if that is what you intended, or else as x^{2(n-1)} (rendered as $x^{2(n-1)}$). Which one is it? –  Harald Hanche-Olsen Jan 28 '13 at 12:22
Thanks, Harald. This is what I was looking for. =) –  Harish Raj Jan 28 '13 at 12:23
Why is this question tagged as "linear-algebra"? –  user1551 Jan 28 '13 at 12:42

Note that $(a^{2^b})^2=a^{2^{b+1}}$.
How can you know whether the OP meant $\,a^{2^b}\,$ or $\,\left(a^2\right)^b=a^{2b}\,$ ? –  DonAntonio Jan 28 '13 at 12:26
$(a^{2^b})^2=a^{2^b}\cdot a^{2^b}=a^{2^b+2^b}=a^{2\cdot2^b}=a^{2^{b+1}}$. –  Harald Hanche-Olsen Jan 28 '13 at 12:26
As @Harald wrote, or slightly more simply: ${\left(a^{2^{b}}\right)}^2 = a^{2 \cdot 2^b} = a^{2^{b+1}}$ –  half-integer fan Jan 28 '13 at 12:28