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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $p$: $$\begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\ & = x^5 + x^4 + (x^4 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + (x^3 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + (x^2 - 1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + x + (x-1) \div (x-1) \\ & = x^5 + x^4 + x^3 + x^2 + x + 1 \end{align*}$$

Answer 4

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

Solution:-

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

Answer 5

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 $a=1,r=x,n=6$.

Answer 6

Do it the hard way with:

Polynomial long division

Answer 7

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.

Answer 8

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 $x.$

Answer 9

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

Answer 10

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

Answer 11

Just apply the following reduction formula inductively.

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

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

Answer 12

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*,

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

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