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.
 A: 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, $$a=b=c=d=e=f=1.$$
A: hint: factor out $(x-1)(1+\cdots x^5)$
A: Hint:- $$y^3-1=y^2(y-1)+y(y-1)+(y-1)=(y-1)\left(y^2+y+1\right)$$
Solution:-

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

A: 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 $p$:
$$\begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\
& = x^5 + x^4 + (x^4 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + (x^3 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + (x^2 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + (x-1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + 1
\end{align*}$$
A: 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 $a=1,r=x,n=6$.
A: 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.
A: Do it the hard way with:
Polynomial long division
A: 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 $x.$
A: Multiply out the right hand side and confirm that the two expressions are equal.
$$\frac{x^6-1}{x-1}=(1+x+x^2+x^3+x^4+x^5)\iff x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)$$
A: 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,
$$f(x)=1+x+x^2+x^3+x^4+x^5.$$
A: Just apply the following reduction formula inductively.

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

Derivation
$$\begin{array}{lll}
\frac{x^n-1}{x-1}&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^{n-1}(x-1)+x^{n-1}-1}{x-1}\\
&=&x^{n-1}+\frac{x^{n-1}-1}{x-1}\\
\end{array}$$
A: 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
$(x-z)(x-z^*)=x^2-(z+z^*)x+zz^*=x^2-2\Re(z)x+|z|^2.$
