For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$ 
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$

My attempt $:$ If $a+b+c \neq 0$ then $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c} = \frac {\left [\{xb+(1-x)c\} + \{xc + (1-x)a\} + \{xa+(1-x)b\} \right]} {a+b+c} =1.$$ Therefore $$x = \frac {a-c}{b-c} = \frac {b-a} {c-a} = \frac {c-b} {a-b}.$$
Comparing the first two expressions of $x$ and simplifying we get \begin{align*} a^2+b^2+c^2 - ab - bc -ca & = 0 \\ \implies (a-b)^2+(b-c)^2 +(c-a)^2 & = 0. \end{align*} Therefore we have $a=b=c.$ But then $x$ would be an indeterminant form. Does it imply that $a+b+c = 0$? How to proceed further? Any help will be highly appreciated. Thank you very much.

 A: if $a=b=c$ is not true one can multiply  three expressions of $x$ in from OP's thisrd step of $x$ to get $x^3=-1$. So $a+b+c \ne 0$ from OP's second equatiom. Hence either $x^3=-1$  or $a=b=c$.
A: $$bc((b-c)x+c)=ca((c-a)x+a)=ab((a-b)x+b)$$
has a solution in $x$ iff
$$\begin{vmatrix}bc(b-c)&bc^2&1\\ca(c-a)&ca^2&1\\ab(a-b)&ab^2&1\end{vmatrix}=-abc(3abc-a^3-b^3-c^3)\\=-abc(a+b+c)(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=0$$
where $\omega$ is a cube root of unity.
There are two cases:

*

*one of the complex factors vanishes when $a=b=c$, and $x$ is indeterminate;


*$a+b+c=0$ and the equations reduce to
$$2x=1.$$
The correct answer is
$$\color{green}{a=b=c\lor \left(a+b+c=0\land x=\frac12\right)}.$$
A: I don't think that's true. Let $x=1/2$. Then:
$$\frac {b+c} {2a} = \frac {c + a}{2b} = \frac {a+b}{2c}.$$
Any $a,b,c\ne 0$ such that $c=-a-b$ and $a+b\ne 0$ are a solution for that:
$$\frac {b-a-b} {2a} = \frac {-a-b + a}{2b} = \frac {a+b}{-2a-2b} \implies \frac {-a} {2a} = \frac {-b}{2b} = \frac {a+b}{-2(a+b)} \implies -\frac {1} {2} = -\frac {1}{2} = -\frac {1}{2}.$$
Moreover, if $a=b=c$, then any value of $x$ satisfies the identity:
$$\frac {xa+(1-x)a} {a} = \frac {xa + (1-x)a}{a} = \frac {xa+(1-x)a}{a} \implies \frac {1} {a} = \frac {1}{a} = \frac {1}{a} \implies 1=1=1.$$
A: A Comment and the full scenario:
$a,b,c,x$ are real and $abc \ne 0$
Case (1) : $a+b+c \ne 0$
(i): $a \ne b \ne c:$
We get $x^3=-1\implies x=-1$ (from OP's work if multiply three expressions of $x$) without $x$ becoming indeterminate ($0/0$).
(i): $a=b=c$ : After OP's second-step we can write two equations for $a,b,c$
as $$-a+xb+(1-x)c=0~~~(1),~~~ (1-x)a-bx+xc=0$$ Solving by Cramer's method, we get
$$\frac{a}{x^2-x+1}=\frac{b}{x^2-x+1}=\frac{c}{x^2-x+1} \implies a=b=c, ~as~ x^2-x-1 \in R, \forall ~x \in R-\{0\}.$$ So when $a+b+c \ne 0$ and $a=b=c$, $x$ is an arbitrary real number.
Case (2): $a+b+c=0:$ The value $x=1/2$ satisfies the the main equation of the question.
Finally we comment that if $a+b+c \ne 0$ and $a=b=c$, then $x=-1$ is only one out of infinitely many real numbers.
