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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.

Thanks in advance.

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    $\begingroup$ yes, I am just looking for a simplification $\endgroup$ May 30, 2013 at 13:08
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    $\begingroup$ 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. $\endgroup$
    – celtschk
    May 30, 2013 at 13:16
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    $\begingroup$ easy on downnvoting gang! this is op's first post $\endgroup$
    – jimjim
    May 30, 2013 at 13:23
  • $\begingroup$ Does this answer your question? Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ $\endgroup$ Aug 17, 2023 at 0:33

3 Answers 3

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If you are looking for this???

$$ a^n-b^n=(a-b)\Big(\sum_{i=0}^{n-1}a^{n-1-i}b^i\Big) $$

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  • $\begingroup$ thank you. What is the name of this? $\endgroup$ May 30, 2013 at 13:12
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    $\begingroup$ By the way it is expansion and not simplification.I dont know a specific name for it. $\endgroup$ May 30, 2013 at 13:13
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$$ a^n \color{#c00}{- b^n} = (a-b+b)^n \color{#c00}{- b^n} = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} \color{#c00}{- b^n} \\= \left[\color{#0c0}{b^n} + {n \choose 1}(a-b)b^{n-1} + {n \choose 2}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n\right] \color{#c00}{- b^n} = (a-b)\left[{n \choose 1}b^{n-1} + {n \choose 2}(a-b)b^{n-2} + {n \choose 3}(a-b)^2b^{n-3} + {n \choose 4}(a-b)^3b^{n-4} +\cdots+ {n \choose n-1} (a-b)^{n-2}b + (a-b)^{n-1} \right] $$ 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.

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$$ a^n -b^n=a^n\Big[1-\Big(\displaystyle\frac{b}{a}\Big)^n\Big] $$ 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: $$ \begin{align} a^n-b^n&=a^n\Big[\Big(1-\frac{b}{a}\Big)\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]=a\cdot a^{n-1}\Big[\Big(1-\frac{b}{a}\Big)\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]\\ &=(a-b)a^{n-1}\Big[\sum_{i=0}^{n-1}\Big(\frac{b}{a}\Big)^i\Big]\\ &=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+ ... +ab^{n-2}+b^{n-1}) \end{align} $$

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