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Math Help - binomial division

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by levan55604 View Post
    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 n\in\mathbb{N}, then:

    \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.
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  3. #3
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    thank you, this is what i was looking for
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  4. #4
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    Hello, levan55604!

    The denominator is (a - b), right?


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

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

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

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

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

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


    Get it?


    Hence: . \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}

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  5. #5
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    As long as a\ne b.
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, levan55604!

    The denominator is (a - b), right?



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

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

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

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

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


    Get it?


    Hence: . \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.
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  7. #7
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    Quote Originally Posted by levan55604 View Post
    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
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