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Math Help - Challenging factorization question?

  1. #1
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    Challenging factorization question?

    If a+b+c = 0 show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)
    Please help.
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  2. #2
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    Re: Challenging factorization question?

    Quote Originally Posted by magakriv View Post
    If a+b+c = 0 show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)
    Please help.
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  3. #3
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    Re: Challenging factorization question?

    Hello, magakriv!

    \text{If }a+b+c \,=\,0,
    . . \,\text{ show that: }\:(2a-b)^3 + (2b-c)^3 + (2c-a)^3 \:=\:3(2a-b)(2b-c)(2c-a)
    Let: . \begin{Bmatrix}X &=& 2a-b \\ Y &=& 2b -c \\ Z &=& 2c-a\end{Bmatrix}

    Note that: / X + Y+Z \:=\:a+b+c \:=\:0


    We have: . X+Y+Z \;=\;0

    Cube both sides: . (X+Y+Z)^3 \;=\;0^3

    . . X^3+Y^3+Z^3+3X^2Y+3XY^2+3Y^2Z+3YZ^2+3Z^2X+3ZX^2+6X  YZ \;=\;0

    X^3+Y^3+Z^3+(3X^2Y+3XY^2+3XY\!Z)+(3Y^2\!Z+3Y\!Z^2+  3XY\!Z)+ (3X^2Z+3XZ^2) \;=\;0

    X^3+Y^3+Z^3 + 3XY(X+Y+Z) + 3YZ(Y+Z+X) + 3XZ(X+Z) \;=\;0


    Since X+Y+Z \,=\,0\,\text{ and }\,X+Z \,=\,-Y, we have:

    . . X^3+Y^3+Z^3 + 3XY(0) + 3Y(0) + 3XZ(-Y) \;=\;0

    . . . . . . X^3+Y^3+Z^3 - 3XYZ \;=\;0

    . . . . . . X^3 + Y^3 + Z^3 \;=\;3XYZ


    Back-substitute: . (2a-b)^3 + (2b-c)^3 + (2c-a)^3 \;=\;3(2a-b)(2b-c)(2c-a)

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  4. #4
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    Re: Challenging factorization question?

    Quote Originally Posted by magakriv View Post
    If a+b+c = 0 show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)
    Please help.
     \text{Let's first consider the cubic expansion of three variables.}

     \begin{aligned} (x+y+z)^3 &= x^3 +y^3 +z^3 +3x^2y+3x^2z + 3y^2x + 3y^2z +3z^2x + 3z^2y + 6xyz\\ &=x^3 + y^3 + z^3 + (3x^2y+3y^2x+3xyz) + (3z^2y+3y^2z+3xyz) + (3z^2x+3x^2z+3xyz) -3xyz\\ &=x^3 + y^3 + z^3 +3xy(x+y+z) +3yz(x+y+z) +3zx(x+y+z) - 3xyz \end{aligned}

     \text{Now let  return to our problem. Given}~ a+b+c=0.}

     \text{Show that}~ (2a-b)^3 + (2b-c)^3 + (2c-a)^3=3(2a-b)(2b-c)(2c-a)

     \text{Let}~ x=2a-b,~ y=2b-c,~ z=2c-a.~ \text{Observe that}~ x+y+z=a+b+c=0

     \text{Substitute}~ x+y+z=0~ \text{into the derivation above, we have}

      0=x^3+y^3+z^3 - 3xyz~ \Leftrightarrow~x^3+y^3+z^3 = 3xyz

     \text{Thus,}~ (2a-b)^3 + (2b-c)^3 + (2c-a)^3=3(2a-b)(2b-c)(2c-a)

    Let me know if you have any question. Good luck!
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  5. #5
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    Re: Challenging factorization question?

    Challenging factorization question?-02-oct-13.png
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