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.
