<s>[INST] 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.

 [/INST] 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.
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