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

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.

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.

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 :-)

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 $X+Y+Z=0$.

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, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$