# Simplifying $\frac{x^6-1}{x-1}$

I have this:

$$\frac{x^6-1}{x-1}$$

I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$

Edit : I was wondering how to do this if I didn't know that it was the same as that.

• What you are looking for is "Polynomial long division." Google it :-) – Eff Dec 7 '14 at 13:52
• It's a pain in the neck to typeset, but long division is the infallible way. Google does turn up a lot of examples. – Cheerful Parsnip Dec 7 '14 at 13:55
• While "polynomial long division" is the way to derive it (i.e. to show this 'if you didn't know that it was the same as that'), this particular result—$\dfrac{x^n-1}{x-1}=1+x+\dotsb+x^{n-1}$—is fairly important, and is worth memorizing. – Akiva Weinberger Dec 7 '14 at 14:23
• Also, if you want to have a better feel for why it's true, try plugging in $x=10$. – Akiva Weinberger Dec 7 '14 at 14:25

Do it the hard way with:

Polynomial long division

• While the information at the link may provide an answer to the question, as it stands this is not an answer. Consider including the essential parts of the answer here, using links only for reference. – user642796 Dec 8 '14 at 5:38

By indeterminate coefficients:

$$(x-1)(ax^5+bx^4+cx^3+dx^2+ex+f)\\=ax^6+(b-a)x^5+(c-b)x^4+(d-c)x^3+(e-d)x^2+(f-e)x-f.$$

After identification, $$a=b=c=d=e=f=1.$$

hint: factor out $(x-1)(1+\cdots x^5)$

• What if I didn't know it could be simplified to that? – raz789 Dec 7 '14 at 13:45
• Then life would be harder (You could just divide them!) – AlonAlon Dec 7 '14 at 14:45

You can easily see that $x^6 - 1$ has a root at $1$, so you know that $$x^6 - 1 = (x-1) \cdot p(x)$$ Where $p(x)$ is a polynomial of degree $5$. Perform polynomial division to find $p$: \begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\ & = x^5 + x^4 + (x^4 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + (x^3 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + (x^2 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + x + (x-1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + x + 1 \end{align*}

Hint:- $$y^3-1=y^2(y-1)+y(y-1)+(y-1)=(y-1)\left(y^2+y+1\right)$$

Solution:-

$y=x^2\implies x^6-1=\left(x^2-1\right)\left(x^4+x^2+1\right)=(x-1)(x+1)\left(x^4+x^2+1\right)$ $$\boxed{\therefore\dfrac{x^6-1}{x-1}=(x+1)\left(x^4+x^2+1\right)=x^5+x^4+x^3+x^2+x+1}$$

• It is interesting that you assume that we can know $x^3-1=(x-1)(x^2+x+1)$ but we cannot know the equivalent for $x^6-1$. – Suzu Hirose Dec 7 '14 at 14:07
• @SuzuHirose: Is it ok now? – user 170039 Dec 7 '14 at 14:45
• Yes I think it's great. – Suzu Hirose Dec 7 '14 at 14:51

If you pay enough attention, you will recognize the sum of a geometric series:

$$\sum_{k=0}^{n-1} ar^k=a\frac{r^n-1}{r-1}.$$ Set $a=1,r=x,n=6$.

Ill show an "tricky" method.

$\displaystyle \frac{x^6 - 1}{x-1}$

$= \displaystyle \frac{x^6 -x + x - 1}{x-1} = \frac{x^6 - x}{x-1} + 1 = \frac{x^6 - x^5 + x^5 - x}{x-1} + 1 = x^5 + 1 + \frac{x^5 - x}{x-1} = \frac{x^5 - x^4 + x^4 - x}{x-1} + x^5 + 1 = \frac{x^4(x - 1) + x^4 - x}{(x-1)}$

Do you see the pattern?

This is simply to show how you can manipulate expressions; its a trick.

here is how i explain this: look at the numbers $9, 99, 999, 9999, \cdots$ in base ten. they are $9 = 10 -1, 99 = (10-1)*11 = 10^2 - 1, 999 = (10-1)*111 = 10^3 - 1, 9999 = (10-1)*1111 = 10^4 - 1$ and the left hand side has the factor $9 = (10-1)$. now you can rewrite the string of equations in the form $$(10 -1) = 1(10 -1),\\ (10^2 - 1) = (10 + 1)(10-1),\\ (10^3 - 1) = (10^2 + 10 + 1)(10-1),\\ (10^4 - 1) = (10^3 + 10^2 + 10 + 1)(10-1), \cdots.$$

now to get your identity think of the polynomials as numbers expressed in base $x,$ that is replace $10$ in the above equations by $x.$

Multiply out the right hand side and confirm that the two expressions are equal. $$\frac{x^6-1}{x-1}=(1+x+x^2+x^3+x^4+x^5)\iff x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)$$

The hard way, by Taylor:

First establish the derivatives, \begin{align} f(x)(x-1)&=x^6-1\\ f'(x)(x-1)+f(x)&=6x^5\\ f''(x)(x-1)+2f'(x)&=30x^4\\ f'''(x)(x-1)+3f''(x)&=120x^3\\ f''''(x)(x-1)+4f'''(x)&=360x^2\\ f'''''(x)(x-1)+5f''''(x)&=720x\\ f''''''(x)(x-1)+6f'''''(x)&=720\\ f'''''''(x)(x-1)+7f''''''(x)&=0\\ \dots\\ f^{(n+1)}(x)(x-1)+(n+1)f^{(n)}(x)&=0\\ \end{align}

Then solve for $x=0$: \begin{align} f(0)&=1,\\ f'(0)=f(0)&=1,\\ f''(0)=2f'(0)&=2,\\ f'''(0)=3f''(0)&=3!\\ f''''(0)=4f'''(0)&=4!\\ f'''''(0)=5f''''(0)&=5!,\\ f''''''(0)=6f'''''(0)-720&=0\\ \dots\\f^{(n)}&=0\end{align} Conclusion, $$f(x)=1+x+x^2+x^3+x^4+x^5.$$

Just apply the following reduction formula inductively.

$$\frac{x^n-1}{x-1} = x^{n-1}+\frac{x^{n-1}-1}{x-1}$$

Derivation $$\begin{array}{lll} \frac{x^n-1}{x-1}&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\ &=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\ &=&\frac{x^{n-1}(x-1)+x^{n-1}-1}{x-1}\\ &=&x^{n-1}+\frac{x^{n-1}-1}{x-1}\\ \end{array}$$

Using complex numbers, you can factor $x^6-1$ from the roots of unit.

$$(x^6-1)=(x-1)(x-e^{i\pi/3})(x-e^{2i\pi/3})(x+1)(x-e^{4i\pi/3})(x-e^{5i\pi/3}).$$ Dropping the factor $(x-1)$ and grouping for the conjugate roots*,

$$(x^2-x+1)(x+1)(x^2+x+1)=(x^4+x^2+1)(x+1)=x^5+x^4+x^3+x^2+x+1.$$

*Using $(x-z)(x-z^*)=x^2-(z+z^*)x+zz^*=x^2-2\Re(z)x+|z|^2.$