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.

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:
$$
\begin{align*}
f(x,y,z) & = xy(az+b) + (x+y)(bz+c) + (cz+d)\\
& = (az+b)[xy + t(x+y) + t^2] 
\end {align*}
$$
