The problem involves a polynomial $F(x, y, z) = x^3 + y^3 + z^3 - 3xyz$. (a) We need to show that $F(x, y, z)$ is a cyclic expression. (b) We need to factorize $F(x, y, z)$. If $F(x, y, z) = 0$ and $(x + y + z) \ne 0$, we need to show that $x^2 + y^2 + z^2 = xy + yz + zx$. (c) If $x = b + c - a$, $y = c + a - b$, and $z = a + b - c$, we need to show that $F(a, b, c) : F(x, y, z) = 1 : 4$.

1. Problem Description

The problem involves a polynomial $F(x, y, z) = x^3 + y^3 + z^3 - 3xyz$.

(a) We need to show that $F(x, y, z)$ is a cyclic expression.

(b) We need to factorize $F(x, y, z)$. If $F(x, y, z) = 0$ and $(x + y + z) \ne 0$, we need to show that $x^2 + y^2 + z^2 = xy + yz + zx$.

(c) If $x = b + c - a$, $y = c + a - b$, and $z = a + b - c$, we need to show that $F(a, b, c) : F(x, y, z) = 1 : 4$.

2. Solution Steps

(a) To show that $F(x, y, z)$ is a cyclic expression, we need to show that $F(y, z, x) = F(x, y, z)$.

$F(y, z, x) = y^3 + z^3 + x^3 - 3yzx = x^3 + y^3 + z^3 - 3xyz = F(x, y, z)$.

Thus, $F(x, y, z)$ is a cyclic expression.

(b) Factorization of $F(x, y, z)$:

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

Given that $F(x, y, z) = 0$ and $x + y + z \ne 0$, we have:

$(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0$.

Since $x + y + z \ne 0$, it implies that $x^2 + y^2 + z^2 - xy - yz - zx = 0$.

Therefore, $x^2 + y^2 + z^2 = xy + yz + zx$.

(c) Given $x = b + c - a$, $y = c + a - b$, and $z = a + b - c$.

$x + y + z = (b + c - a) + (c + a - b) + (a + b - c) = a + b + c$.

$F(x, y, z) = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.

$F(a, b, c) = a^3 + b^3 + c^3 - 3abc$.

$F(a, b, c) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Let's calculate $F(x, y, z)$ with given $x, y, z$.

$x + y + z = a + b + c$.

Now, we calculate $x^2 + y^2 + z^2 - xy - yz - zx$.

$x^2 + y^2 + z^2 = (b + c - a)^2 + (c + a - b)^2 + (a + b - c)^2$

$= b^2 + c^2 + a^2 + 2bc - 2ab - 2ac + c^2 + a^2 + b^2 + 2ca - 2bc - 2ab + a^2 + b^2 + c^2 + 2ab - 2ac - 2bc$

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

$xy + yz + zx = (b + c - a)(c + a - b) + (c + a - b)(a + b - c) + (a + b - c)(b + c - a)$

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

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

$xy + yz + zx = (b+c-a)(c+a-b) + (c+a-b)(a+b-c) + (a+b-c)(b+c-a) = -a^2 - b^2 - c^2 + 2ab + 2bc + 2ca = -(a^2 + b^2 + c^2) + 2(ab+bc+ca)$

$x^2 + y^2 + z^2 - (xy+yz+zx) = 3(a^2 + b^2 + c^2) - 2(ab + bc + ca) - [-(a^2 + b^2 + c^2) + 2(ab+bc+ca)] = 4(a^2 + b^2 + c^2) - 4(ab + bc + ca) = 4[a^2 + b^2 + c^2 - (ab + bc + ca)]$.

$F(x, y, z) = (a + b + c) [4(a^2 + b^2 + c^2 - ab - bc - ca)] = 4(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

$F(a, b, c) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

$F(a, b, c) / F(x, y, z) = [(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)] / [4(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)] = 1 / 4$.

Therefore, $F(a, b, c) : F(x, y, z) = 1 : 4$.

3. Final Answer

(a) $F(x, y, z)$ is a cyclic expression.

(b) $F(x, y, z) = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$. If $F(x, y, z) = 0$ and $(x + y + z) \ne 0$, then $x^2 + y^2 + z^2 = xy + yz + zx$.

(c) $F(a, b, c) : F(x, y, z) = 1 : 4$.