# 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. – Sharkos Oct 29 '13 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. – Seirios Oct 29 '13 at 15:58
• @Seirios: Don't both do so? Glad we agree on the relevant thing anyway :) – Sharkos Oct 29 '13 at 21:49

## 12 Answers

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. – Marc van Leeuwen Jun 7 '13 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. – Marc van Leeuwen Jun 7 '13 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, $b,c$. – Sharkos Jun 7 '13 at 19:13

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! – Jyrki Lahtonen Oct 29 '13 at 16:05

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! – Jyrki Lahtonen Jun 7 '13 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, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

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 '13 at 13:00
• Relating the question to Newton's identities is really illuminating: bravo! – Georges Elencwajg Oct 28 '13 at 8:25

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? – lab bhattacharjee Jun 7 '13 at 11:28

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?” – Michael Hoppe Oct 28 '13 at 17:28