8
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ yes, I am just looking for a simplification $\endgroup$ May 30 '13 at 13:08
  • 1
    $\begingroup$ 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 '13 at 13:16
  • 1
    $\begingroup$ easy on downnvoting gang! this is op's first post $\endgroup$
    – jimjim
    May 30 '13 at 13:23
8
$\begingroup$

If you are looking for this???

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

$\endgroup$
2
  • $\begingroup$ thank you. What is the name of this? $\endgroup$ May 30 '13 at 13:12
  • 1
    $\begingroup$ By the way it is expansion and not simplification.I dont know a specific name for it. $\endgroup$ May 30 '13 at 13:13
7
$\begingroup$

$$ a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} - b^n = (a-b)b^{n-1} + {n \choose 1}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n = (a-b)(b^{n-1} + {n \choose 1}(a-b)b^{n-2} + {n \choose 2}(a-b)^2b^{n-3} + {n \choose 3}(a-b)^3b^{n-4} + ... + {n \choose n-1} (a-b)^{n-2}b + (a-b)^{n-1} ) $$ 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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.