Results 1 to 4 of 4

Thread: (a + b + c)^3

  1. #1
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448

    (a + b + c)^3

    Verify this identity: (a + b + c)^3 + (a - b - c)^3 - (a + c - b)^3 - (a + b - c)^3 = 24abc
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    This is just a matter of expanding each bracketed expression, and then simplifying everything. Tedious, more than anything. Here's the first bracketed term expanded.

    $\displaystyle (a + b + c)^3$

    $\displaystyle =(a+b+c)(a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2)$

    $\displaystyle =(a+b+c)(a^2 + 2ab + 2ac + b^2 + 2bc + c^2)$

    $\displaystyle =a^3+3a^2 b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3b c^2+c^3$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello pacman
    Quote Originally Posted by pacman View Post
    Verify this identity: (a + b + c)^3 + (a - b - c)^3 - (a + c - b)^3 - (a + b - c)^3 = 24abc
    First, can I encourage you to use LaTeX: it makes your questions and working much more readable. In this instance, all you need do to change your original expression to LaTeX is to select it, and then press the $\displaystyle \Sigma$ button on the toolbar. Which is what I've done here:

    $\displaystyle (a + b + c)^3 + (a - b - c)^3 - (a + c - b)^3 - (a + b - c)^3 = 24abc$

    Then: I think we can reduce the tedium a bit by noting one or two well-known results first:

    $\displaystyle p^3 + q^3 = (p+q)(p^2-pq+q^2)$ (1)

    $\displaystyle p^2-q^2 = (p-q)(p+q)$ (2)

    $\displaystyle (p+q)^2+(p-q)^2 = 2p^2+2q^2$ (3)

    We begin by re-grouping a little:

    $\displaystyle (a+b+c)^3+(a-b-c)^3-(a+c-b)^3 -(a+b-c)^3 $$\displaystyle =\Big((a+[b+c])^3+(a-[b+c])^3\Big)-\Big((a+[c-b])^3 +(a-[c-b])^3\Big)$

    Then, using (1) on each expression in the big brackets:

    $\displaystyle = 2a\Big((a+[b+c])^2-(a+[b+c])(a-[b+c]) +(a-[b+c])^2\Big)$ - $\displaystyle 2a\Big((a+[c-b])^2-(a+[c-b])(a-[c-b]) +(a-[c-b])^2\Big)$

    $\displaystyle =2a\Big(2a^2 +2(b+c)^2 -a^2 +(b+c)^2\Big)-2a\Big(2a^2+2(c-b)^2-a^2+(c-b)^2\Big)$, using (2) and (3)

    $\displaystyle =2a\Big(a^2 +3(b+c)^2\Big)-2a\Big(a^2+3(c-b)^2\Big)$

    $\displaystyle =2a\Big(3(b+c)^2-3(c-b)^2\Big)$

    $\displaystyle =6a(b^2+2bc+c^2-c^2+2bc-b^2)$

    $\displaystyle =24abc$

    Grandad
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    Thanks Grandad, your method is excellent . . . . i have tried same as that of Rowe, it overlapped my paper. . . . .

    And i will learn LateXing . . . .
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum