20201015, 10:01  #12 
Romulan Interpreter
Jun 2011
Thailand
5^{2}·7·53 Posts 
Or look for decomposition of \(a^n\pm b^n\) for \(n\) odd or even.
For why \(x^{ab}=(x^a)^b\) think about definition of powers, as a repeated multiplication. To get \(x^n\) you multiply \(x\) by itself \(n\) times. Write down \(x\) \(n\) times, and group it in \(b\) groups of \(a\) items each. Now you have \(x^a\), written \(b\) times. Last fiddled with by LaurV on 20201015 at 10:10 
20201015, 14:04  #13  
Feb 2017
Nowhere
2^{4}·271 Posts 
Quote:
x^{a} * x^{b} = x^{a+b} (x^{a})^{b} = x^{a*b} The first law is initially proven for positive integers a and b; repeated application gives the second law. Both laws can then be extended to all integer exponents by demanding that these laws continue to hold. In particular, x^{0} is 1. The fact that x  y is an algebraic factor of x^{n}  y^{n} may be proved by mathematical induction, as indicated in this post, with a simplified argument in this post. Substituting x^{a} for x and y^{a} for y, and using the second law of exponents above, then tells you that if b is odd, x^{a} + y^{a} is an algebraic factor of x^{ab} + y^{ab} 

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