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 March 5th, 2013, 03:41 AM #1 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Factorization Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$ Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$ Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - \{(c+a)^3 +(a+b)^3 +(b+c)^3\}$ Putting (c+a) =A ; (a+b) =B ; (b+c) =C $\{(c+a)^3 +(a+b)^3 +(b+c)^3\}$ becomes $(A^3+B^3+C^3)$now again using the formulae : $a^3+b^3+c^3= (a+b+c)^3-3(a+b)(b+c)(c+a)$ Could you please help me in this if there is any other easier way of getting it .....
 March 5th, 2013, 04:40 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Factorization It's not that complicated as it looks like : $(a + b + c)^3 - (b + c)^3 - (c + a)^3 - (a + b)^3 + a^3 + b^3 + c^3$ $= (a + b + c)^3 - 2(a^3 + b^3 + c^3) - 3ab(a + b) - 3bc (b + c) - 3ac (a + c) + a^3 + b^3 + c^3$ $= (a + b + c)^3 - a^3 - b^3 - c^3 - 3ab(a + b) - 3bc (b + c) - 3ac (a + c)$ $= 3(a + b)(b + c)(c + a) - 3ab(a + b) - 3bc (b + c) - 3ac (a + c)$ $= 3abc + 3abc = 6abc$ And the result follows.

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