Find the value of $(a^3 + b^3 + c^3)/(abc)$ if $a/b + b/c + c/a = 1$. 
Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$

I tried using Cauchy's inequality but it was of no help. Please guide me. 
$a, b, c$ are real.
 A: There is not enough information to solve this problem. Clearing out denominators, your hypothesis is
$$a^2 c + a b^2 + b c^2 = abc \quad (1)$$
and your desired conclusion is
$$a^3+b^3+c^3=kabc \quad (2)$$
for some constant $k$. 
Suppose, for the sake of contradiction, there were a $k$ such that $(1)$ implied $(2)$. Since the polynomial $a^3+b^3+c^3-abc$ is irreducible, this would mean that  $a^2 c+a b^2+b c^2-abc$ would divide $a^3+b^3+c^3-kabc$. But the two polynomials are both cubics, so the only way for the first to divide the second is the first is a scalar multiple of the second, and it isn't.
It's also easy to generate points on $(1)$ and see that the ratio $(a^3+b^3+c^3)/(abc)$ isn't constant. Just choose random values for $a$ and $b$ and equation $(1)$ turns into a quadratic; solving that quadratic gives you some points to try. You'll see very quickly that nothing like this is true.
A: Let 
$$\begin{eqnarray}
u&=&a+b+c\\
v&=&a^3+b^3+c^3\\
s&=&a^2b + b^2 c + c^2 a\\
t &=& a \, b\, c
\end{eqnarray}$$
These are related by 
$$u^3 = v + 3 s + t $$
Further, we are given
$$ \frac ab + \frac bc + \frac ca = \frac st =1$$
so $$u^3 = v + 4 t$$
And our target is
$$\frac{a^3+b^3+c^3}{abc}=\frac vt = \frac{u^3}{t}-4$$
This cannot be a single value, as we have a remaining degree of freedom. 
A: Take $a=1$ and $b=-\epsilon$. Then $c=O(1/\epsilon)$ ($\epsilon\rightarrow 0$). By taking $\epsilon$ suitably small, we can make the expression to be evaluated, of order $1/\epsilon^3$, big or much bigger.
