<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] $$bc((b-c)x+c)=ca((c-a)x+a)=ab((a-b)x+b)$$
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
$$\color{green}{a=b=c\lor \left(a+b+c=0\land x=\frac12\right)}.$$
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