Value of $\frac{x^n -a^n}{ x-a}$? Is there a rule that governs the value of $\frac{x^n-a^n}{x-a}$? For instance $\frac{x^{12}-1}{x-1}$ is equal to $x^{11} + x^{10}... +1$. How is this result obtained? There is no limit over here, so I don't think there is calculus used, also I'm still in a beginner stage in calculus.
 A: The correct formula is, for $x\ne y$:
$$\frac{x^n-y^n}{x-y}=x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1}=\sum_{i=0}^{n-1} x^{n-1-i}y^{i}$$
You prove that by expanding the product:
$$(x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})$$
When $y=1$ and $x\ne1$, you get indeed
$$\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}\dots+x+1$$
Advice: never forget the additional condition that $x\ne y$. You will often come upon a sum like $1+x+\dots+x^{n-1}$, and it's always tempting to blindly simplify to $\dfrac{x^n-1}{x-1}$, but when $x=1$, the sum is equal to $n$, while the fraction is not defined.
A: For fun, an exercise in long division.
Consider $p(x):= x^n -a^n.$
$p(a)=0$, hence $x-a$ is a factor.
By polynomial long division:
${\tiny (x^n-a^n)÷(x-a)= x^{n-1} + ax^{n-2}+a^2x^{n-3}.....+a^{n-1}}$
${\tiny -(x^n -ax^{n-1})}$

${\tiny (ax^{n-1}-a^n)}$
${\tiny -(ax^{n-1}-a^2x^{n-2})}$

${\tiny a^2x^{n-2} -a^n}$
${\tiny -(a^2x^{n-2} - a^3x^{n-3})}$

:
:
${\tiny a^{n-1}x-a^n}$
${\tiny -(a^{n-1}x -a^n)}$

${\tiny 0}$
A: Calculus is only required for the limit (if any) as the values of $n$ get infinitely large.  
For any finite positive integer value of $n$ then (assuming $x \ne a$ then
$\frac {x^n - a^n}{x-a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}=\sum\limits_{k=0}^{n-1} x^ka^{(n-1) - k}$.
The reason why is that if
$(x-a)K = x^n - x^a$ and if $x\ne a$ then $K = \frac {x^n-x^a}{x-a}$ and:
$(x-a)( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) = $
$x( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) - $
$a( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) =$
$(x^n + x^{n-1}a + ......+x^2a^{n-2} +xa^{n-1}) -$
$(x^{n-1}a + x^{n-2}a^2 + .......... + xa^{n-1} + a^n)=$
$x^n + (x^{n-1}a - x^{n-1}a) + (x^{n-2}a^2 -x^{n-2}a^2) + ..... + (xa^{n-1} - xa^{n-1}) - a^n = $
$x^n - a^n$.
(This is called "telescoping" as it "folds up like a telescope";  when you multiply one term $x^ia^j$ by $x$ to get $x^{i+1}a^j$ and you multiply the next term $x^{i+1}a^{j-1}$ by $-a$ you get $-x^{i+1}a^j$ and they cancel each other out.  And only the very first term, $x^{n-1}$ times $x$ and the very last term $a^{n-1}$ times $-a$ are left.)
This is a handy and very well known "trick".  Look at it.  Get it.  Chuckle at it because it is so clever.  And never forget it.  It will be very useful later and you will see it a lot[*].  
.......
In calculus we figure out what happens when 
$n\to \infty$ and $\lim\limits_{n\to \infty} \frac {x^n - a^n}{x-a} = \lim\limits_{n\to \infty}(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1})$
and we can use it too figure out that.
$1 + \frac 12 + (\frac 12)^2 + (\frac 12)^3+ ......... = $
$\lim \frac {1^n - (\frac 12)^n}{1 - \frac 12} = \frac {1 - 0}{\frac 12} = 2$
And $1 + \frac 23 + (\frac 23)^2 + (\frac 23)^3+ ......... = $
$\lim \frac {1^n - (\frac 23)^n}{1 - \frac 23} = \frac {1 - 0}{\frac 23} = 1\frac 13$
And than if $|r| < 1$ then $ 1 + r + r^2 + ..... = \lim \frac{1-r^n}{1-r} = \frac 1{1-r}$.
Which is ... kind of neat.
But that's calculus and you'll learn that later.
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[*]And don't forget when in a year or two a novice asks you about it, be sure to look down at them with a "well, isn't obvious" look and show the trick and make sure they feel like crap for not coming up with themselves .... and suggest they maybe would be better off dropping mathematics altogether and taking up something a little more "creative" like basket weaving.
This is an important aspect of math learning... Not sure, why but it seems to be.
