# How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly?

I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?

• A general formula for $A^n+B^n+C^n$ in terms of the symmetric polynomials $$s_1=A+B+C, s_2=AB+BC+AC, s_3=ABC$$ is $$A^n+B^n+C^n = \sum_{i+2j+3k=n} (-1)^j\frac{n}{i+j+k}\binom{i+j+k}{i,j,k} s_1^is_2^js_3^k$$ Not sure how that would help, but the RHS has a term $+3s_3$ when $n=3$. – Thomas Andrews Aug 24 '13 at 22:31
• Specifically, $$A^3+B^3+C^3 = (A+B+C)^3 - 3(A+B+C)(AB+BC+AC) + 3ABC$$ – Thomas Andrews Aug 24 '13 at 22:35

## 6 Answers

We know $(A+B)\mid(A^n+B^n)$ for $n$ odd. What about with three terms? Compute

$$\mod A+B+C:\quad A^3+B^3+C^3\equiv-(B+C)^3+B^3+C^3\equiv-3BC(B+C)\equiv 3ABC.$$

So $(A+B+C)\mid(A^3+B^3+C^3-3ABC)=f(A,B,C)$. Further

$$f(A,B,C)=f(A,\omega B,\bar{\omega}C)=f(A,\bar{\omega}B,\omega C)=\rm etc.$$

by inspection so both $A+\omega B+\bar{\omega}C$ and $A+\bar{\omega}B+\omega C$ are also factors.

This argument exploits divisibility properties and inherent symmetry. It is generalized by the first proof (using matrix operations) mentioned in my other answer to compute $\Phi(G)$ for $G$ abelian.

HINT: If $A+Bw+Cw^2=0$ where $w$ is one of the three cube roots of unity

$\implies -A=Bw+Cw^2$

Cubing we get, $(-A)^3=(Bw+Cw^2)^3$

$\implies -A^3=B^3w^3+C^3w^6+3\cdot Bw\cdot Cw^2(Bw+Cw^2)=B^3+C^3+3BC(-A)$

$\implies A+Bw+Cw^2$ is a factor of $A^3+B^3+C^3-3ABC$

You can consider it as a polynomial in $A$ and attempt to factor it. So you want to find polynomials in $B$ and $C$, say $r,s,t$, such that

$$(A + r)(A+s)(A+t)=A^3-A(3BC)+B^3+C^3.$$

In particular, you need $r+s+t=0$, and you similarly have information about $rs+st+tr$ and $rst$. It's not hard to see that the roots are linear polynomials in $B$ and $C$, so they have the form $a+bB+cC$ for a constant $a$. You can plug this representation into the three equations you got from looking at the coefficients and solve.

If you need help getting started on the resulting equations, note that $rst$ has no constant term, so at least one of the roots has no constant term.

This is a $3\times3$ circulant determinant which is a special case of a group determinant ($G=C_3$).

Let $\{X_g\}$ be a set of formal variables indexed by elements of a group $G$, then $\Phi(G):=\det(X_{gh^{-1}})$ we define to be the group determinant. KCd has a set of notes concerning these objects in the history of representation theory, and includes two proofs of the factorization

$$\Phi(G)=\prod_{\chi\in\widehat{G}}\left[\sum_{g\in G}\chi(g)X_g\right]$$

for finite abelian groups $G$. The first proof shows each linear factor divides $\Phi(G)$ by invoking matrix row operations, the other exhibits the factors as eigenvalues of a linear transformation.

More generally invoking the Wedderburn decomposition yields for arbitrary finite $G$

$$\Phi(G)=\prod_{\rho~\rm irred}\det\left(\sum_{g\in G}X_g\rho(g)\right)^{\deg\rho}.$$

Expanding on Potato's answer:

First lets make $a^3 + b^3 + c^3 -3abc\$ a polynomial in $a$ so we get $a^3 - a3bc + b^3 + c^3$. This has $3$ roots so lets express it as $(a+p)(a+q)(a+r)$. But also know: $$a^3 + b^3 + c^3 -3abc = (a+b+c)(a^2 + b^2 + c^2 -ab -bc - ac)$$ Therefore lets say $p=b+c$. So our polynomial is now: $(a+q)(a+r)(a+b+c)$. Expanding this expression and equating the co-efficients of $a$ we get the following equations: $$r+q + b+c = 0$$ $$qr+br+bq+cr+cq = -3bc$$ $$bqr + cqr = b^3 + c^3$$ Doing sum of roots on the third equation we can get: $$qr = b^2 - bc +c^2$$ subbing the this equation into the second equation we get: $$(b+c)^2 + br+bq+cr+cq = 0$$ $$(b+c)^2 = -(b+c)(q+r)$$ $$b+c = -(q+r)$$ Now, since $b+c$ is real and lets assume $q$ and $r$ are complex, this means $r$ is the conjugate of $q$. So let $q=x+yi$. Therefore we get: $$-2x = b+c$$ $$qr = |q|^2 = x^2+y^2 = b^2+c^2-bc$$ So we have $x = -(b+c)/2$ and substituting this in the above equation we get $y = \frac{\sqrt{3}}{2}(b-c)$.

This means: $$q= -\frac{b+c}{2} + \frac{\sqrt{3}}{2}(b-c)i$$ $$= -\frac{b}{2} + \frac{\sqrt{3}}{2}bi + -\frac{c}{2} - \frac{\sqrt{3}}{2}ci$$ $$= bw^2 + cw$$ where w is the complex cubic root of unity. Hence this means that: $$r = cw^2 + bw$$ Therefore: $$a^3 + b^3 + c^3 -3abc = (a+b+c)(a+ bw + cw^2)(a+bw^2 + cw)$$

$(a^3+b^3+c^3) = (a+b+c)(a^2 +b^2+c^2-ab-bc-ac)$

$1+\omega + \omega^2 = 0 \implies \omega+\omega^2 = -1$

$= (a+b+c)(a^2 + b^2 + c^2 +ab(\omega+\omega^2)+bc(\omega+\omega^2)+ac(\omega+\omega^2)) = (a+b+c)(a(a+b\omega^2+c\omega)+b\omega(a+b\omega^2+ c\omega)+ c\omega^2 (a+ \dfrac b\omega+c\omega)$

Now, note that $\dfrac{1}{\omega}= \omega^2$

And after substituting this we get the desired identity.