Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers 
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but second is not an extreme value. Comment on particular cases when (i) ${a+b+c=0}$, (ii) ${a=b=c}$.

My Attempt:
$${F=f+\lambda \phi =x^3 + y^3 + z^3 - 3xyz + 3\lambda(ax + by + cz-1)}$$
$${\frac13F_x = x^2-yz+\lambda a = 0}{\text{ ...(1)}}$$
$${\frac13F_y = y^2-xz+\lambda b = 0}{\text{ ...(2)}}$$
$${\frac13F_z = z^2-xy+\lambda c = 0{\text{ ...(3)}}}$$
$${(1)x+(2)y+(3)z \implies f+\lambda (1) = 0 \implies \lambda = -f}$$
$${(1)+(2)+(3) \implies}{x^2+y^2+z^2-xy-yz-zx=(a+b+c)f}$$ 
$${\implies f/(x+y+z)=(a+b+c)f}\,,$$ 
then ${f=0}$ or ${x+y+z=1/(a+b+c)}$.
Also, ${(1)-(2) \implies x^2-y^2-z(x-y)=f(a-b) \implies \frac{x-y}{a-b}=f\frac{a+b+c}{x+y+z}}$
Similarly ${(2)-(3)}$ and ${(3)-(1)}$, then we get
${\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=f\frac{a+b+c}{x+y+z}}$
I don't know how to proceed further. I couldn't get the other stationary value of ${f}$. I need help in proceeding further to calculate stationary points and/or stationary values, if at all, my method is correct.

Without Lagrange Multipliers.
Making the change of variables $y = \lambda x, z = \mu x$ and substituting we get
$$
\min\max x^3(1+\lambda^3+\mu^3-3\lambda\mu)\ \ \mbox{s. t. }\ \ x(a+\lambda b+\mu c) = 1
$$
or equivalently
$$
\min\max f(\lambda,\mu) = \frac{1+\lambda^3+\mu^3-3\lambda\mu}{(a+\lambda b+\mu c)^3}
$$
whose stationary points are solved by
$$
\left\{
\begin{array}{rcl}
 \left(\lambda ^2-\mu \right) (a+c \mu )-b \left(\mu ^3-2 \lambda  \mu +1\right)& = & 0 \\
 c \left(\lambda ^3-2 \mu  \lambda +1\right)+(a+b \lambda ) \left(\lambda -\mu ^2\right) & = & 0 \\
\end{array}
\right.
$$
giving the points
$$
\begin{array}{ccc}
\lambda & \mu & f(\lambda,\mu)\\
 1 & 1 & 0 \\
 \frac{b^2-a c}{a^2-b c} & \frac{c^2-a b}{a^2-b c} & \frac{1}{a^3+b^3+c^3-3 a b c} \\
\end{array}
$$
