# 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

• @Seirios It's worth noticing that even the most trivial of edits (making that formula display style rather than inline) pushes a question to the front page. Now whilst I enjoy the extra rep, there's arguably a lot of wasted effort on behalf of all the people writing new answers. In essence, I suggest people avoid necroing for trivial formatting. Oct 29, 2013 at 15:23
• @Sharkos: Strictly speaking, Jyrki Lahtonen's bounty pushed the question to the front page and not really my edit. Nevertheless I agree with you. Oct 29, 2013 at 15:58
• @Seirios: Don't both do so? Glad we agree on the relevant thing anyway :) Oct 29, 2013 at 21:49

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 delightful way of looking at it! Oct 29, 2013 at 16:05

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.

• I suppose that the "sum of cubes" referred to is $(a-b)^3 + (b-c)^3 + (c-a)^3$. Since it is homogeneous of degree $3$ I cannot see how you can call it quadratic. True, none the the variables has degree $3$ individually, but that is not what is usually called quadratic. Jun 7, 2013 at 15:24
• Indeed I don't see what the first part is used for at all; the "sum of cubes" is homogeneous of degree $3$ and divisible by $(a-b)(a-c)(b-c)$, and that already makes it a scalar multiple of the latter. Jun 7, 2013 at 15:26
• Yes that is what I referred to. It is quadratic in each variable individually. This just allows you to deduce the form of the dependence on one variable, say $a$, just by computing two roots, $b,c$. Jun 7, 2013 at 19:13

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 :-)

• Yes, sometimes swatting flies with cannonballs is fun! Jun 7, 2013 at 19:17

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

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 $x=a-b,y=b-c,z=c-a$

• @user42912, do you call it a calculation? Jun 7, 2013 at 11:28

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 $x_1=a-b$, $x_2=b-a$, $x_3=c-a$.

• This is essentially the calculation made by @lab bhattacharjee.
– lhf
Jun 7, 2013 at 13:00
• Relating the question to Newton's identities is really illuminating: bravo! Oct 28, 2013 at 8:25

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!

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

$$g(b)=g(0)=g(2b)=g(-b)=0 \Rightarrow g(x)\equiv 0$$

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 $P/Q=\frac{-3}{-1}=3$.

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 $0$

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

• Done “without calculations?” Oct 28, 2013 at 17:28