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  1. #1
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    algebra

    if a+b+c=0, prove (2a-b)^3 +(2b-c)^3 +(2c-a)^3
    = 3 (2a-b)(2b-c)(2c-a)
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  2. #2
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    Quote Originally Posted by Frank Chan View Post
    if a+b+c=0, prove (2a-b)^3 +(2b-c)^3 +(2c-a)^3
    = 3 (2a-b)(2b-c)(2c-a)
    Do you remember this formula


    x^3 +y^3 +z^3 - 3xyz = (x+y+z)(x^2 +y^2 +z^2 - xy-yz-za)
    We will use it

    when x+y+z=0

    Then x^3+y^3+z^3 = 3xyz
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Now in LHS
    Consider

    2a-b = x

    2b -c =y

    2c-a= z

    Thus x+y+z =
    (2a - b) + (2b - c) +(2c -a) = 2(a+b+c) - (a+b+c) = 0

    -------------
    Hence
    x^3 + y^3 +z^3 = (2a-b)^3 +(2b-c)^3 +(2c-a)^3

    But using formula

    = 3xyz = 3 (2a-b)(2b-c)(2c-a) =RHS

    Thus proved
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  3. #3
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    Hello, Frank!

    If a+b+c\:=\:0, prove: . (2a-b)^3 +(2b-c)^3 +(2c-a)^3 \:=\: 3 (2a-b)(2b-c)(2c-a)

    Let: . \begin{array}{cccc}X &=& 2a-b & {\color{blue}[1]} \\ Y &=& 2b-c & {\color{blue}[2]} \\ Z &=& 2c-a & {\color{blue}[3]} \end{array}

    We want to prove that: . X^3 + Y^3 + Z^3 \:=\:3XYZ



    Add [1], [2], [3]: . X + Y + Z \:=\:a+b+c \quad\Rightarrow\quad X+Y+Z \:=\:0 .[4]

    . . and we have: . \begin{array}{ccc} X+Y &=& \text{-}Z \\ X+Z &=& \text{-}Y \\ Y+Z &=& \text{-}X\end{array} .[5]


    Cube both sides of [4]: . (X+Y+Z)^3 \:=\:0^3

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

    . . X^3 + Y^3 + Z^3 + 3XY(X+Y) + 3YZ(Y+Z) + 3XZ(X+Z) + 6XYZ \;=\;0


    Substitute [5]: . X^3 + Y^3 + Z^3 + 3XY(\text{-}Z) + 3YZ(\text{-}X) + 3XZ(\text{-}Y) + 6XYZ \;=\;0

    . . . . . . . . X^3 + Y^3 + Z^3 - 3XYZ - 3XYZ - 3XYZ + 6XYZ \;=\;0

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


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


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  4. #4
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    Thumbs up Thank you

    Thank you for your help.

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