-1

I am a bit lost how to factor $a^{k} - b^{k}$. I know it links to the binomial theorem but I can't remember how to do it. Could anyone explain?

Hoping_Blessing
  • 417
Andrew
  • 245
  • From question 712758:

    $Q=\sum_{i=1}^{n-1}{a^{n-i}b^i}=a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}$

    $a^n-b^n=(a-b)Q=(a)Q+Q(-b)$

    $\begin{matrix} (a)Q&=&a^n&+a^{n-1}b&+a^{n-2}b^2&+\dots&+a^2b^{n-2}&+ab^{n-1}&\ +Q(-b)&=&&-a^{n-1}b&-a^{n-2}b^2&-\dots&-a^2b^{n-2}&-ab^{n-1}&-b^n\ \end{matrix}$

     – Senex Ægypti Parvi Jun 07 '15 at 22:18

2 Answers2

6

It has nothing to see with the binomial theorem, but with the geometric series. Actually it's a high-school formula: $$1+x+x^2+\dots+x^{k-1}=\frac{1-x^k}{1-x},$$ which also reads as $$1-x^k=(1-x)(1+x+x^2+\dots+x^{k-1})$$ whence, settink $x=\dfrac ba$, and multiplying both sides by $a^k$, one obtains: $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+a^{k-3}b^2+\dots+b^{k-1}).$$

Bernard
  • 175,478
0

We have

$$\begin{align}a^k-b^k&=(a-b)\left(a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1}\right)\end{align}$$

marwalix
  • 16,773