<s>[INST] Help solve this through complex numbers I have to do the following problem:

If $ax+cy+bz = X, cx+by+az=Y, bx+ay+cz=Z$, then show that
$(a^2+b^2+c^2-bc-ca-ab)(x^2+y^2+z^2-yz-zx-xy) = X^2+Y^2+Z^2-YZ-XZ-XY$.

It's easy enough through normal Algebra if I pair $X^2-XY = X(X-Y)$ etc., and collect the coefficients on both sides, but the chapter is of complex numbers and more specifically, cube roots of unity. How do I solve this using that idea?
The best I've come up with is to replace $-1$ with $\omega+\omega^2$, but I later seem to apply the reverse and thus, render the use of $\omega$ pointless. Some guidance, please!
 [/INST] HINT:
$$\sum_{\text{cyc}}(X^2-YZ)=X^2+Y^2+Z^2+(XY+YZ+ZX)(w^2+w)$$
$$=(X+Yw+Zw^2)(X+Yw^2+Zw)$$ where $w$ is a complex cube root of $1$
$$X+Yw+Zw^2=ax+cy+bz +w(cx+by+az)+w^2(bx+ay+cz)$$
$$=a(x+w^2y+wz)+bw^2(x+y^2w+wz)+cw(x+y^2w+wz)=(x+y^2w+wz)(a+bw^2+cw)$$
$$\text{Similarly, for  } X+Yw^2+Zw$$

Alternatively, without using Complex Numbers,
$$X^2-YZ=(ax+cy+bz)^2-(cx+by+az)(bx+ay+cz)$$
$$=(a^2-bc)x^2+(c^2-ab)y^2+(b^2-ca)z^2+(2ca-b^2-ca)xy+(2ab-c^2-bc)zx+(2bc-ab-a^2)yz$$
$$\implies\sum_{\text{cyc}}(X^2-YZ)=(x^2+y^2+z^2-xy-yz-zx)(a^2+b^2+c^2-ab-bc-ca)$$
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