<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] 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:
$$\frac {xa+(1-x)a} {a} = \frac {xa + (1-x)a}{a} = \frac {xa+(1-x)a}{a} \implies \frac {1} {a} = \frac {1}{a} = \frac {1}{a} \implies 1=1=1.$$
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