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Without expanding either side, show that.\((a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a-b)(b-c)(c-a)\) I have read proofs saying that you can  Let a−b=x,b−c=y,c−a=z.a−b=x,b−c=y,c−a=z and prove it but I don't really understand can somebody please show me how to show that it's equal without expanding the equation? A step by step approach would really help, thank you!!!

 May 28, 2018

Prove: \((a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)\)


Using the method your proofs say, here is what they mean. 


Basically, the proof is assigning each expression another varible. 


\(a-b=x\\ b-c=y\\ c-a=z\\\)

So instead of a - b, b - c, or c - a, they replace these expressions with x, y, and z. 


When we do replace them with x, y, and z, we get:




Since \(x+y+z=a-b+b-c+c-a=0\), we get \(x + y +z = 0\)


Using that, we get:






Therefore, \((a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)\)


If hope this helped,



 May 28, 2018
edited by GYanggg  May 28, 2018
edited by GYanggg  May 28, 2018

I have noticed a flaw in my proof. 


One may not understand why \(x^3+y^3+z^3−3xyz=(x+y+z)(x^2+y^2+z^2−xy−xz−yz)=0\)


Here is why: 


Using \(a^3 + b^3 = (a + b)(a^2 – ab + b^2)\)


We get: \(x^3+y^3+z^3−3xyz=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz\)

Since I added and subtracted \(3x^y\ \text{and}\ 3xy^2\), the value of the expression does not change. 

I did this so I can factor out some values: 


\(\boxed{x^3+y^3+z^3−3xyz\\ =x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3x^2y-3xyz\\ =(x+y)^3+z^3-3xy(x+y+z)\\ =(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)\\ =(x+y+z)(x^2+y^2+z^2-xy-xz-yx)}\)


If you still don't understand something, private message me and I will try to resolve the issue. 



 May 28, 2018
edited by GYanggg  May 28, 2018
edited by GYanggg  May 29, 2018
edited by GYanggg  May 29, 2018
edited by GYanggg  May 29, 2018

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