How to simplify $a^n - b^n$?

How to simplify $a^n - b^n$?

If it would be $(a+b)^n$, then I could use the binomial theorem, but it's a bit different, and I have no idea how to solve it.

• yes, I am just looking for a simplification May 30 '13 at 13:08
• I'd say it is already the absolute simplest form I can think of. Just as $(a+b)^n$ is much simpler than what you get after using the binomial theorem. May 30 '13 at 13:16
• easy on downnvoting gang! this is op's first post May 30 '13 at 13:23

$$a^n-b^n=(a-b)\Big(\sum_{i=0}^{n-1}a^{n-1-i}b^i\Big)$$
$$a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} - b^n = (a-b)b^{n-1} + {n \choose 1}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n = (a-b)(b^{n-1} + {n \choose 1}(a-b)b^{n-2} + {n \choose 2}(a-b)^2b^{n-3} + {n \choose 3}(a-b)^3b^{n-4} + ... + {n \choose n-1} (a-b)^{n-2}b + (a-b)^{n-1} )$$ too complicated to simplify if you use bionomial expansion.
$$a^n -b^n = a^n -a^{n-1}b + a^{n-1}b -b^n = a^{n-1}(a-b) + b(a^{n-1}-b^{n-1}) = a^{n-1}(a-b) + b(a^{n-1} -a^{n-2}b + a^{n-2}b -b^{n-1}) = a^{n-1}(a-b) + b(a^{n-2}(a-b) + b(a^{n-2}-b^{n-2})) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2(a^{n-2}-b^{n-2}) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2 a^{n-3}(a-b) + b^3(a^{n-3}-b^{n-3}) = ...(repeat ) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2 a^{n-3}(a-b) + b^3 a^{n-4}(a-b) + ... + b^{n-2}a(a-b) + b^{n-1}(a-b) = (a-b)(a^{n-1} + b a^{n-2} + b^2 a^{n-3} + ... + b^{n-2}a + b^{n-1})$$ the power of $a$ and $b$ add up to $n-1$ in the second factor.
$$a^n -b^n=a^n\Big[1-\Big(\displaystyle\frac{b}{a}\Big)^n\Big]$$ If $a=b$, then $a^n -b^n$ is equal to zero and there is no need for a formula. Suppose $a\neq b$ and $\displaystyle\frac{a}{b}\neq 1$. Then we can use the formula for the computation of a finite geometric series which is: $$\sum_{i=0}^{n-1}r^i=\frac{1-r^n}{1-r},\quad r\neq 1$$ Let $r=\displaystyle\frac{b}{a}$ in the above formula. This gives us: \begin{align} a^n-b^n&=a^n\Big[\Big(1-\frac{b}{a}\Big)\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]=a\cdot a^{n-1}\Big[\Big(1-\frac{b}{a}\Big)\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]\\ &=(a-b)a^{n-1}\Big[\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]\\ &=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+ ... +ab^{n-2}+b^{n-1}) \end{align}