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

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?

  • 1
    $\begingroup$ 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... $\endgroup$ – DonAntonio Jan 28 '13 at 12:05
  • $\begingroup$ oh, sure. I was wondering the same how I could do that. $\endgroup$ – Harish Raj Jan 28 '13 at 12:06
  • $\begingroup$ 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? $\endgroup$ – Harald Hanche-Olsen Jan 28 '13 at 12:22
  • $\begingroup$ Thanks, Harald. This is what I was looking for. =) $\endgroup$ – Harish Raj Jan 28 '13 at 12:23
  • $\begingroup$ Why is this question tagged as "linear-algebra"? $\endgroup$ – user1551 Jan 28 '13 at 12:42

Note that $(a^{2^b})^2=a^{2^{b+1}}$.

  • $\begingroup$ Gerry, Can you explain me how? (sorry, I am weak in math) $\endgroup$ – Harish Raj Jan 28 '13 at 12:25
  • $\begingroup$ How can you know whether the OP meant $\,a^{2^b}\,$ or $\,\left(a^2\right)^b=a^{2b}\,$ ? $\endgroup$ – DonAntonio Jan 28 '13 at 12:26
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    $\begingroup$ $(a^{2^b})^2=a^{2^b}\cdot a^{2^b}=a^{2^b+2^b}=a^{2\cdot2^b}=a^{2^{b+1}}$. $\endgroup$ – Harald Hanche-Olsen Jan 28 '13 at 12:26
  • $\begingroup$ @DonAntonio He couldn't, but it was a (very) reasonable guess. $\endgroup$ – Harald Hanche-Olsen Jan 28 '13 at 12:27
  • $\begingroup$ As @Harald wrote, or slightly more simply: ${\left(a^{2^{b}}\right)}^2 = a^{2 \cdot 2^b} = a^{2^{b+1}}$ $\endgroup$ – half-integer fan Jan 28 '13 at 12:28

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