# Thread: binomial division

1. ## binomial division

can someone please show me how to solve this: (a^n-b^n)/(a-n), and i would also like to know if there's more than one way to do it. Thank you

2. Originally Posted by levan55604
can someone please show me how to solve this: (a^n-b^n)/(a-n), and i would also like to know if there's more than one way to do it. Thank you
If $\displaystyle n\in\mathbb{N}$, then:

$\displaystyle \displaystyle \frac{a^n-b^n}{a-b} = \frac{\left(a-b\right)\sum_{k=0}^{n-1}a^{n-k-1}b^k}{a-b} = \sum_{k=0}^{n-1}a^{n-k-1}b^k.$

3. thank you, this is what i was looking for

4. Hello, levan55604!

The denominator is $\displaystyle (a - b)$, right?

Can someone please show me how to solve this? . $\displaystyle \dfrac{a^n-b^n}{a-b}$

Use long division on the first few values of $\displaystyle \,n$ and note the pattern.

$\displaystyle \dfrac{a^2-b^2}{a-b} \;=\;a + b$

$\displaystyle \dfrac{a^3-b^3}{a-b} \;=\; a^2 + ab + b^2$

$\displaystyle \dfrac{a^4-b^4}{a-b} \;=\; a^3 + a^2b + ab^2 + b^3$

$\displaystyle \dfrac{a^5 - b^5}{a-b} \;=\;a^4 + a^4b + a^2b^2 + ab^3 + b^4$

Get it?

Hence: .$\displaystyle \dfrac{a^n-b^n}{a-b} \;=\;a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \hdots + a^2b^{n-3} + ab^{n-2} + b^{n-1}$

5. As long as $\displaystyle a\ne b$.

6. Originally Posted by Soroban
Hello, levan55604!

The denominator is $\displaystyle (a - b)$, right?

Use long division on the first few values of $\displaystyle \,n$ and note the pattern.

$\displaystyle \dfrac{a^2-b^2}{a-b} \;=\;a + b$

$\displaystyle \dfrac{a^3-b^3}{a-b} \;=\; a^2 + ab + b^2$

$\displaystyle \dfrac{a^4-b^4}{a-b} \;=\; a^3 + a^2b + ab^2 + b^3$

$\displaystyle \dfrac{a^5 - b^5}{a-b} \;=\;a^4 + a^4b + a^2b^2 + ab^3 + b^4$

Get it?

Hence: .$\displaystyle \dfrac{a^n-b^n}{a-b} \;=\;a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \hdots + a^2b^{n-3} + ab^{n-2} + b^{n-1}$

Thanks for your reply. I'm having trouble trying to figure out how to solve this problem using the long division. I'm feeling like an idiot , can you please show me how to do it. If there's only one variable I can do it in seconds without the calculator but add a second variable and I'm lost.

7. Originally Posted by levan55604
Thanks for your reply. I'm having trouble trying to figure out how to solve this problem using the long division. I'm feeling like an idiot , can you please show me how to do it. If there's only one variable I can do it in seconds without the calculator but add a second variable and I'm lost.
I figured it out. Here's how to do it: for example if we have (27a^3-8b^3)/(3a-2b)

9a^2
__________________________________________________ ___
3a-2b| 27a^3-8b^3

27a^3 /3a =9a^2
------------------------------------------------------------------
------------------------------------------------------------------

9a^2
__________________________________________________ __
3a-2b| 27a^3-8b^3
27a^3 -18ba^2
__________________________________________________ __
-8b^3 +18ba^2

(9a^2) (3a) =27a^3, (9a^2)(-2b) = -18ba^2

------------------------------------------------------------------
------------------------------------------------------------------

9a^2 +4b^2
__________________________________________________ __
3a-2b| 27a^3-8b^3
27a^3 -18ba^2
__________________________________________________ __
-8b^3 +18ba^2

(-8b^3)/(-2b)= 4b^2

------------------------------------------------------------------
------------------------------------------------------------------

9a^2 +4b^2
__________________________________________________ __
3a-2b| 27a^3-8b^3
27a^3 -18ba^2
______________________________________________
-8b^3 +18ba^2
-8b^3 +12ab^2
_______________________________________
(18ba^2)-(12ab^2)

(4b^2)(-2b)= -8b^3, (4b^2)(3a)= 12ab^2

------------------------------------------------------------------
------------------------------------------------------------------

9a^2 +4b^2 +6ab
__________________________________________________ __
3a-2b| 27a^3-8b^3
27a^3 -18ba^2
______________________________________________
-8b^3 +18ba^2
-8b^3 +12ab^2
_______________________________________
(18ba^2)-(12ab^2)

(18ba^2)/ (3a) = 6ab

------------------------------------------------------------------
------------------------------------------------------------------

9a^2 +4b^2 +6ab
__________________________________________________ __
3a-2b| 27a^3-8b^3
27a^3 -18ba^2
______________________________________________
-8b^3 +18ba^2
-8b^3 +12ab^2
_______________________________________
(18ba^2)-(12ab^2)
18ba^2 -12ab^2
_____________________

(6ab)(3a)= 18ba^2, (6ab)(-2b) = -12ab^2

so the answer is : 9a^2 +6ab+4b^2