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Math Help - Another factoring problem

  1. #1
    Junior Member DIOGYK's Avatar
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    Another factoring problem

    Hello, it's me again with my factoring problems....

    Problem 122. Factor:
    (c) a^3+b^3+c^3-3abc;

    My attempt: Only thing I was able to do is to expand a^3+b^3=a^3+a^2b-a^2b+ab^2-ab^2+b^3

    Also, please explain the solution step by step.

    Thanks in advance.
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  2. #2
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    Re: Another factoring problem

    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
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  3. #3
    Junior Member DIOGYK's Avatar
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    Re: Another factoring problem

    Quote Originally Posted by zaidalyafey View Post
    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
    Thanks for the reply, but I already found that on other sites, how you get that is my question.
    Last edited by DIOGYK; August 21st 2012 at 12:39 AM.
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    Re: Another factoring problem

    a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc

    add and subtract the terms \,\,a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\,\, then use grouping

    a^3+a^2b+a^2c+b^3+b^2a+b^2c+c^3+c^2a+c^2b-abc-a^2b-ab^2-abc-c^2b-b^2c-abc-a^2c-ac^2

    a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c)-bc(a+b+c)-ac(a+b+c)

    (a+b+c)(a^2+b^2+c^2-ab-ac-bc)
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  5. #5
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    Re: Another factoring problem

    a^3+b^3=(a+b)(a^2-ab+b^2 )
    Sub b=b+c
    a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 )
    a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )
    a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c)
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc)
    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)



    This is one method of getting that hope you understand...
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  6. #6
    Junior Member DIOGYK's Avatar
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    Re: Another factoring problem

    Quote Originally Posted by zaidalyafey View Post
    a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc

    add and subtract the terms \,\,a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\,\, then use grouping

    a^3+a^2b+a^2c+b^3+b^2a+b^2c+c^3+c^2a+c^2b-abc-a^2b-ab^2-abc-c^2b-b^2c-abc-a^2c-ac^2

    a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c)-bc(a+b+c)-ac(a+b+c)

    (a+b+c)(a^2+b^2+c^2-ab-ac-bc)
    Thank you for the explanation! That was really helpful!
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  7. #7
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    Re: Another factoring problem

    on thing to notice about a^3 + b^3 + c^3 - 3abc is that it is symmetric in a,b and c: switch any two of a,b, and c and you still get the same thing.

    so if it factors, it will factor as two symmetric expressions, one involving just a single term at a time, and one involving two terms at a time.

    but the only symmetric expression in a,b,c involving just a single term at a time is: a+b+c, or some multiple thereof. so let's divide a^3 + b^3 + c^3 - 3abc by a+b+c,

    considering it as a polynomial in a.

    obviously the first term of the quotient is going to be a^2, so we get:

    (a + b + c)(a^2) = a^3 + a^2b + a^2c. subtracting that from a^3 + b^3 + c^3 - 3abc gives us:

    -(b+c)a^2 - (3bc)a + b^3 + c^3. the a^2 term's coefficient is -(b+c), so we take:

    (a+b+c)(-(b+c)a) = -(b+c)a^2 - (b^2 + bc)a - (bc + c^2)a

     = -(b+c)a^2 - (b^2 + c^2 + 2bc)a, and subtract THAT from our previous remainder:

    -(b+c)a^2 - (3bc)a + b^3 + c^3 -(-(b+c)a^2 - (b^2 + c^2 + 2bc)a)

    = -(b+c)a^2 - (3bc)a + b^3 + c^3 + (b+c)a^2 + (b^2 + c^2 + 2bc)a

     = (b^2 + c^2 - bc)a + b^3 + c^3.

    now the coefficient of a is b^2 + c^2 - bc, so we take:

    (a+b+c)(b^2 + c^2 - bc) = (b^2 + c^2 - bc)a + b^3 + bc^2 - b^2c + b^2c + c^3 - bc^2

     = (b^2 + c^2 - bc)a + b^3 + c^3.

    as this is equal to our last remainder, we have:

    a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 -(b+c)a + b^2 + c^2 - bc)

    or, re-arranging a little:

    a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc).
    Last edited by Deveno; August 21st 2012 at 02:51 AM.
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  8. #8
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    Re: Another factoring problem

    Another approach:

    a^3 + b^3 + c^3 -3abc = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} = \begin{vmatrix}a+b+c & b & c \\ a+b+c & a & b \\ a+b+c & c & a \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{vmatrix} = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ac)

    Source: "My Favourite Polynomial", Desmond MacHale, The Mathematical Gazette, Vol. 75, No. 472 (June 1991), pp. 157-165.
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  9. #9
    Junior Member DIOGYK's Avatar
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    Re: Another factoring problem

    Quote Originally Posted by sujith123 View Post
    a^3+b^3=(a+b)(a^2-ab+b^2 )
    Sub b=b+c
    a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 )
    a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )
    a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c)
    a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc)
    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)



    This is one method of getting that hope you understand...
    Thank you, your explanation was the easiest to understand.
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