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!!!

Guest May 28, 2018

#1**+1 **

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:

\(x^3+y^3+z^3=3(x)(y)(z)\)

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

Using that, we get:

\(x^3+y^3+z^3−3xyz=(x+y+z)(x^2+y^2+z^2−xy−xz−yz)=0\)

\(a-b=0\Rightarrow{a}=b\)

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

If hope this helped,

Gavin.

GYanggg
May 28, 2018

#2**+2 **

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.

Gavin

GYanggg
May 28, 2018