-1

If $(a+b+c)(x+y+z)=p+q+r$ then prove that $$(a^3+b^3+c^3 - 3abc)(x^3+y^3+z^3 -3xyz)=p^3+q^3+r^3 -3pqr$$

I proceeded in the following way:

Suppose $A=a+b+c$ and $B= x+y+z$. Then $AB=p+q+r$. Now $$\begin{align} \text{L.H.S.} &= (a^3+b^3+c^3 - 3abc)(x^3+y^3+z^3 -3xyz) \tag1\\[4pt] &= (a+b+c)\left((a+b+c)^3-3(ab-bc-ca)\right) \\ &\quad\cdot(x+y+z)\left((x+y+z)^3-3(xy-yz-zx)\right) \tag2\\[4pt] &= (p+q+r)\left(A^3-3(ab-bc-ca)\right)\left(B^3-3(xy-yz-zx)\right) \tag3 \end{align}$$ Then I got the trouble.

Blue
  • 83,939
VINI
  • 1
  • 2
    A well known factorisation might help here: $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = \frac{1}{2} (a+b+c)\left[(a-b)^2 + (b-c)^2 + (c-a)^2 \right]$$ – Kraken Nov 19 '24 at 06:06
  • @Kraken I think it more likely that the full factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a+\omega b+\omega^2c)(a+\omega^2 b+\omega c)$$ with $\omega=e^{2\pi i/3}$ a complex third root of unity will play a role. – Jyrki Lahtonen Nov 19 '24 at 07:19
  • See this old thread. The answers by Anon in particular. – Jyrki Lahtonen Nov 19 '24 at 07:40

2 Answers2

3

I don't think what you want to prove is true:

Let $a=x=1,b=y=2,c=z=3$. Then $$(a^3+b^3+c^3 - 3abc)(x^3+y^3+z^3 -3xyz) = 18^2=324.$$

Let $p=10,q=12,r=14$, then $p+q+r=36=(a+b+c)(x+y+z)$ but $$p^3+q^3+r^3 -3pqr=432$$

Tri
  • 1,513
  • 1
    Well spotted. On the other hand, if you pick $p=10, q=r=13$, then the identity does hold. I will type an explanation shortly. – Jyrki Lahtonen Nov 19 '24 at 07:24
3

Warning: Not a full answer, but too long to be a comment. As Tri observed, this does not always hold. My point is that we do get this identity if we specify $p,q,r$ a bit more carefully.

I'm fond of handling this three variable cubic polynomial as follows.

Consider the circulant matrix $$ M(a,b,c)=\left(\begin{array}{ccc} a& b& c\\ c&a&b\\ b&c&a\end{array}\right). $$ It is well known that we get the cubic as the determinant $$ F(a,b,c):=\det M(a,b,c)=a^3+b^3+c^3-3abc. $$

What the question reminds anyone well versed in linear algebra is the determinant product rule $$ \det A \cdot \det X= \det AX. $$ We obviously want $X=M(x,y,z)$. Furthermore, it is known that the product of two circulant matrices is itself circulant. Indeed, $$ M(a,b,c) M(x,y,z)= M(a x + c y + b z, b x + a y + c z, c x + b y + a z). $$ Putting all this together implies that $$ F(a,b,c) F(x,y,z)=F(a x + c y + b z, b x + a y + c z, c x + b y + a z). $$ In other words, the claim holds, if we, instead, specify $$ \begin{cases} p=ax+cy+bz,\\ q=bx+ay+cz,\\ r=cx+by+az. \end{cases} $$

The claim does hold more generally. For example, we can obviously permute $\{p,q,r\}$ any way we want. I'm afraid I cannot tell, what would be a sufficient and necessary condition for the identity to hold. The given constraint is satisfied with the above choice of $p,q,r$, but is not sufficient.

For the benefit of readers familiar with basic abstract algebra concepts: the algebra of $3\times 3$ circulant matrices is isomorphic to that of the ring $K[x]/\langle x^3-1\rangle$ where $K$ is the field we draw the constants from (most likely $K=\Bbb{R}$ or $K=\Bbb{C}$). The matrix $M(a,b,c)$ represents multiplication by the coset of $a+bx+cx^2$ w.r.t. the basis consisting of the cosets of $\{1,x,x^2\}$.

Jyrki Lahtonen
  • 140,891