Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ is a geometric progression. 
Let $a,b,c,d$ be real numbers not all $0$, and let $f(x,y,z)$ be the  polynomial in three variables deﬁned by
  $$f(x,y,z) = axyz + b(xy + yz + zx) + c(x + y + z) + d.$$
Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ is a geometric progression.

What I have tried...
Let $g$ is symmetric  We may assume that $g = x + y + z + k$ for some constant $k.$
And I'm stuck here!!, any help will be appreciated.
 A: 
It's nice question, and little hard too, and I think it's Olympiad question, I wounder from what Olympiad did you take this question?, This is my solution :

Suppose that $f$ is reducible. Therefore it has a factor $g$ of degree $1$. 
Suppose that $g$ is symmetric. We may assume that $$g = x + y + z + k$$ for some constant $k.$
Now put $x = 0$, so  $ y + z + k$ must divide $byz + e$ for some constant $e$.
However, any product of $my + nz + t$ with $y + z + k$ will involve either $y^{2}$ or $z^{2}$ terms unless $m = n = 0$, so this is impossible.
On the other hand, if $g$ is not symmetric, say $g = ux + vy + wz + k$ where $u,v,w$ are not all equal. Then by symmetry, $f$ is divisible by
$$(ux + vy + wz + k)(vx + wy + uz + k)(wx + uy + vy + k).$$
Now by equating coeﬃcients of $x^{3}$ and of $x^{2}y$ we ﬁnd that either: $u = v = 0$, $v = w = 0$ or $w = u = 0$ and these give rise to the terms $a,b,c$ and $d$ being in geometric progression because $$f = a(x + r)(y + r)(z + r).$$
This argument relies upon the polynomials in question forming part of a Unique Factorization Domain, and in fact this is true since $\mathbb{R}[x,y,z]$ is a UFD since we wish to conclude that $f$ is divisible by the displayed polynomial on the basis that it is divisible by each of its three given factors.
This factorization result (or assumption) is often overlooked in elementary algebra texts, and we do not propose to punish students who make this implicit assumption. However, there is a way to do this problem by a direct calculation which avoids the issue.
Observe that $f = Q(ux + vy + wz + k) + R$ where $Q = ayz + by + bz + c$ and $$R = byz + cy + cz + d−(αx + βy + γ)(ayz + by + bz + c)$$ which is identically $0$. The coeﬃcients of $R$ must all be $0$. Equate coeﬃcients:
$y^{2}z , yz^{2}    : −aα = −aβ = 0 \tag{1}$
$y^{2} , z^{2} : −bα = −bβ = 0 \tag{2}$
$yz : −aγ −bα−bβ + b = 0 \tag{3}$
$y, z : c−cα−bγ = 0,c−cβ −bγ = 0 \tag{4}$
$1 : d−cγ = 0 \tag{5}$
If $a = b = 0$, then f is linear, and hence irreducible. Thus we assume that $a = 0$ or $b = 0.$
Then it follows from $(1)$ and $(2)$ that $α = β = 0.$
It then follows from $(3)$ that $b = aγ,$ and from $(4)$ that $c = bγ,$ and from $(5)$ that $d = cγ.$
Thus $(a;b;c;d) = (a;aγ;aγ^{2};aγ^{3}) $ as required.
A: We have:
$$
\begin{align*}
f(x,y,z) & = axyz + b(xy + yz + zx) + c(x + y + z) + d.\\
& = axyz + b(xy) + b(yz + zx) + c(x+y)+ cz + d\\
& = xy(az+b) + (x+y)(bz+c) + (cz+d)\\ 
\end {align*}
$$
The only way we can make it reduce further is if $(az+b), (bz+c)$ and $(cz+d)$ are related geometrically, i.e., $(bz+c)$ and $(cz+d)$ are of the type $m\cdot(az+b)$.
Let the geometric progression be $a, b=at, c=at^2, d=at^3$.  
Then $(bz+c) = at(z) + at^2 = t(az+at) = t(az+b)$. Similarly, $(cz+d) = t^2(az+b)$.
Now, back to our equation:
$$
\begin{align*}
f(x,y,z) & = xy(az+b) + (x+y)(bz+c) + (cz+d)\\
& = (az+b)[xy + t(x+y) + t^2] 
\end {align*}
$$
A: First note that we can assume that $a=1$ so we have $f(x,y,z)=xyz+b(xy+yz+zx)+c(x+y+z)+d$.
We have that $xyz+r(xy+yz+zx)+r^2(x+y+z)+r^3=(x+r)(y+r)(z+r)$ for the if part.
Noe looking at this as an expression polynomial in $x$ it has degree $1$. Because we can invert scalars any factorisation will be of the form $$B(x+A)$$ where $A,B$ are polynomial in $y,z$.
Since $f(x,y,z)=(yz+b(y+z)+c)x + byz+c(y+z)+d$ we must have $B=yz+b(y+z)+c$ whence 
$Ayz+Ab(y+z)+Ac=byz+c(y+z)+d$ and from this we get $A=b$, $Ab=c=b^2$ and $Ac=d=b^3$
