How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
 A: This is the two-variable identity $(X+Y)^3 - (X^3 + Y^3) = 3XY(X+Y)$  
presented as a formula for $X^3 + Y^3 + Z^3$ when $X+Y+Z=0$  (by setting $Z = -(X+Y)$),   
and then using $(X,Y,Z)=(a-b,b-c,c-a)$ to parametrize solutions of $X+Y+Z=0$.
A: Let $x_1$, $x_2$, $x_3$ be such that $x_1+x_2+x_3=0$. Using Newton's identities, we have $ p_3 = e_1p_2 - e_2p_1 + 3e_3$, where $p_k$ is the sum of the $k$-th powers of the $x_i$ and $e_k$ is the $k$-th  elementary symmetric polynomial in the $x_i$. Since $e_1=p_1=0$ by hypothesis, we have $p_3=3e_3$.
The result in the question follows by taking $x_1=a-b$, $x_2=b-a$, $x_3=c-a$.
A: Notice that this is invarient if the same constant is added to all three variables, so one of them may be set equal to zero. Set $c=0$. Then the expression becomes $(a-b)^3 -a^3 +b^3+3ab(a-b)$, easily seen to be zero.
   BTW, apparently the original expression is equivalent to saying $x^3 +y^3+z^3 =3xyz$ if $x+y+z=0$.
EDIT:Yes, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$
A: We know $$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$
If $x+y+z=0, x^3+y^3+z^3=-3(x+y)(y+z)(z+x)=-3(-z)(-x)(-y)=3xyz$
Alternatively if $x+y+z=0,$
$x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3=(-z)^3-3xy(-z)+z^3=3xyz$
Put $x=a-b,y=b-c,z=c-a$
A: An equivalent identity is that $x+y+z = 0$ implies $x^3+y^3+z^3-3xyz = 0$.
So suppose $x+y+z = 0$.
Then the determinant $ \begin{vmatrix} x & z & y\\ y & x & z\\ z & y & x \end{vmatrix}$ must be zero, because the sum of the elements of each column is zero.
Expanding the determinant, we have the required result.
A: We use the identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
let $x=a-b ,y=b-c, z=c-a$  we see $x+y+z=0$ so $(a-b)^3+(b-c)^3+(c-a)^3-3(a-b)(b-c)(c-a)=0 $
Done!
A: Let $$f(x)=(x-b)^3 + (b-c)^3 + (c-x)^3 -3(x-b)(b-c)(c-x) $$
The $f$ is a polynomial of degree at most $3$. Moreover, $x^3$ appears twice, with coefficients $\pm 1$, thus cancel. It follows that $f$ is at most quadratic and it is obvious that
$$f(c)=f(b)=f(0)=0$$
Since $f$ is at most quadratic and has three roots, $f \equiv 0$.
Note: If you want to avoid the observation that the coefficients of $x^3$ cancel, then use $f(c)=f(b)=f(0)=0$ and
$$f(b+c)=c^3+(b-c)^3-b^3+3c(b-c)(-b)=c^3-b^3+(b-c)^3-3bc(b-c)$$
You can see that this is zero, or if you still want to avoid this computation, repeat the above argument:
Let $g(x)=x^3-b^3+(b-x)^3-3bx(b-x)$. Then $g$ is at most cubic and 
$$g(b)=g(0)=g(2b)=g(-b)=0 \Rightarrow g(x)\equiv 0$$
A: The term $P=(a-b)^3 + (b-c)^3 + (c-a)^3$ is cyclically symmetric in $a,b,c$, and divisible by $a-b$ since setting $a=b$ makes it zero. It is therefore divisible by
$Q=(a-b)(b-c)(c-a)$ (whose factors are pairwise relatively prime), and given that $\deg(P)=3=\deg(Q)$, the quotient must be a constant. Computing the coefficients of $a^2b$ in $P,Q$ (admittedly a computation) one sees that $P/Q=\frac{-3}{-1}=3$.
A: The easiest way to show this is by observing that the sum of cubes is such that all the cubed terms cancel, so it is quadratic in each variable individually; then notice that the sum of cubes vanishes for $a=b,c$ and for $b=c$. Consequently it must factorize as $(a-b)(a-c)(b-c)\times d$ for some $d$. (Why? One gets $f(b,c)\times (a-b)(a-c)$ by thinking of it in terms of a quadratic in $a$; then the form of $f$ follows by thinking in terms of $b$ or simply symmetry.) Letting $a,b,c=0,1,2$ tells you the constant.
Alternatively, note that $(a-b)$ must be a factor, so by cyclic symmetry $(a-b)(b-c)(c-a)$ must be. The result can be deduced similarly from the above.
A: You can also do the following. Let us replace $a$ with $x$ and treat the l.h.s. as a polynomial function $f(x)$ of $x$. Let's check the derivative
$$
f'(x)=3(x-b)^2-3(c-x)^2-3(b-c)(c-x)+3(x-b)(b-c).
$$
Continuing (we could see that this is identically zero, but I try to avoid such manipulations) we see
$$
f''(x)=6(x-b)+6(c-x)+3(b-c)+3(b-c)=0
$$
for all $x$. As $f'(c)=3(c-b)^2+0+0+3(c-b)(b-c)=0$, we can then conclude that $f'(x)=0$ for all $x$, so $f(x)$ is a constant.
But
$$f(b)=0+(b-c)^3+(c-b)^3+3\cdot0=0,$$
so the claim follows.

The OP also asked for an argument using some concepts from abstract algebra.
Consider the function
$$
f(x,y,z)=(x-y)^3+(y-z)^3+(z-x)^3+3(x-y)(y-z)(z-x).
$$
We can easily check that under permutations of the variables the polynomial $f$ changes its sign according to the parity of the permutation:
$$
-f(y,x,z)=-f(x,z,y)=-f(z,y,x)=f(x,y,z)=f(y,z,x)=f(z,x,y).
$$
The space of homogeneous polynomials of degree three in the three variables is a vector space of dimension ten. It is easy to calculate the character $\chi$ of this representation of the symmetric group $G=S_3$ using the basis of monomials.
We get $\chi(1_G)=10$ and $\chi(x\mapsto y\mapsto z\mapsto x)=1$ as $xyz$ is the only monomial stable under a 3-cycle. We get $\chi(x\mapsto y\mapsto x, z\mapsto z)=2$ as the monomials $xyz$ and $z^3$ are both invariant under this substitution. Let $\sigma$ be the sign character.
We can compute their inner product
$$
\langle \chi,\sigma\rangle=\frac16(1\cdot10-3\cdot2+2\cdot)=\frac66=1.
$$
Thus our 10-dimensional space $V$ has only a 1-dimensional subspace $W$ transforming according to the sign character under permutation of the variables.
But clearly the polynomials $(x-y)(y-z)(z-x)$ and $(x-y)^3+(y-z)^3+(z-x)^3$ both transform like that, i.e. belong to $W$. Hence they must be scalar multiples of each other. Fixing $z$ and $y$ and computing the limit
$$
\lim_{x\to\infty}\frac{(x-y)(y-z)(z-x)}{(x-y)^3+(y-z)^3+(z-x)^3}=\frac13
$$
then gives the conclusion :-)
A: Let $ a-b = x $ , $ b -c= y $ and $  c-a =z$ . So actually you want to know the value of  $ x^3+y^3+z^3-3xyz$ . Here the technique resides . 
$ x+y = (a-c) =-z $ ,
$x^3+y^3+z^3-3xyz = (x+y)^3-3xy(x+y) +z^3-3xyz
                   = -z^3 +z^3 +3xyz -3xyz  = 0 $
Hence the value is $ 0 $
A: To Prove:
$$(a-b)^3 + (b-c)^3 + (c-a)^3 =3(a-b)(b-c)(c-a)$$
we know, $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
so, $$(a-b)^3 + (b-c)^3 = (a -c)((a-b)^2 - (a-b)(b-c) + (b-c)^2)$$
now, $$(a-b)^3 + (b-c)^3 + (c-a)^3 = (a -c)((a-b)^2 - (a-b)(b-c) + (b-c)^2) + (c-a)^3 =
(c-a)(-(a-b)^2 + (a-b)(b-c)- (b-c)^2 +(c-a)^2)$$
now, $(c-a)^2 - (a-b)^2 = (c-a+a-b)(c-a-a+b) = (c-b)(c-2a+b)$
the expression becomes,
$$(c-a)((c-b)(c-2a+b) + (b-c)(a-2b+c)) = (c-a)(b-c)(-c+2a-b+a-2b+c)=3(c-a)(b-c)(a-b)$$
Hence proved
