Suppose $a\not=b \not=c$. \[ a(a^2+b+c) - b(b^2+c+a) = 0 \]\[ a^3 + ac - b^3 - bc = 0 \]\[ (a-b)(a^2+ab+b^2 + c) = 0\]Therefore, $a^2 + ab + b^2 + c =0$ Do this similarly to the other equations, we'll get \[ a^2 + ab + b^2 + c = 0 \]\[ b^2 + bc + c^2 + a = 0\]\[ c^2 + ac + a^2 + b = 0 \]\[ a^2 + ab + b^2 + c = b^2 + bc + c^2 + a \]\[ a^2 - c^2 + ab - bc + c - a = 0 \]\[ (a - c)(a + c + b - 1) = 0 \]Therefore, we know that $a + b + c = 1$. Now, \[ a(a^2+b+c) = 1 \]\[ a(a^2 + 1 - a) = 1 \]\[ a^3 + a - a^2 - 1 = 0 \]\[ (a - 1)(a^2 + 1) = 0 \]Note that $a,b,c$ satisfy $x(x^2+1-x) = 1$ simultaneously. Therefore, $a = b = c = 1$ which obviously doesn't satisfy. I don't know but it's quite weird. Please correct me.

$c$ can be negative as well.

ultralako wrote: $c$ can be negative as well. Oh yeah sorry. I misread the problem.

sqing wrote: Let $a, b, c$ be non-zero real numbers such that $a^2+b+c=\frac{1}{a}, b^2+c+a=\frac{1}{b}, c^2+a+b=\frac{1}{c}.$ Prove that at least two of $a, b, c$ are equal. Solution. To seek a contradiction, assume that $a\ne b$, $b\ne c$, and $a\ne c$. Thus \begin{align*}&a^2+b+c=\frac{1}{a}\text{ and }b^2+c+a=\frac{1}{b}\\ \Longrightarrow&(a^2+b+c)-(b^2+c+a)=\frac{1}{a}-\frac{1}{b}\\ \Longrightarrow&(a-b)(a+b-1)=\frac{b-a}{ab}\\ (\because a-b\ne0)\Longrightarrow&a+b+\frac{1}{ab}=1. \end{align*}Likewise, $a^2+b+c=\frac{1}{a}$ and $c^2+a+b=\frac{1}{c}$ gives $a+c+\frac{1}{ac}=1$. Hence$$a+b+\frac{1}{ab}=1=a+c+\frac{1}{ac}\Longrightarrow b-c=\frac{b-c}{abc}\Longrightarrow abc=1,$$which implies $a+b+c=a+b+\frac{1}{ab}=1$, which, combined with $a^2+b+c=\frac{1}{a}$, gives $a^2+1-a=\frac{1}{a}$, namely, $a^3-a^2+a-1=0$, i.e., $(a-1)(a^2+1)=0$ thereby $a=1$. Likewise, we obtain $b=1$, contradicting the assumption $a\ne b$. $\blacksquare$

sqing wrote: Let $a, b, c$ be non-zero real numbers such that $a^2+b+c=\frac{1}{a}, b^2+c+a=\frac{1}{b}, c^2+a+b=\frac{1}{c}.$ Prove that at least two of $a, b, c$ are equal. Do you know the name of the author for this problem @sqing sir ?

So here goes my solution; FTSOC.$a\neq b\neq c$, $a^2+b+c=\frac{1}{a}$, we multiply this by $a$ and we get $a^3+ab+ac=1$ similarly for the others we get $b^3+bc+ab=1$ and $c^3+ac+bc=1$ Combining the first 2 we get that $a^3-b^3=c(b-a)$, since we supposed that $a\neq b\neq c$ we can divide by $(a-b)$ and we get that $a^2+ab+b^2=-c$ similarly with others we get $a^2+ac+c^2=-b$ $b^2+bc+c^2=-a$ summign 2 of them we get $(b^2-a^2)+c(b-a)=b-a$ , again since $a\neq b\neq c$ we can divide by $b-a$ We get that $a+b+c=1$ Now we get back to the first statement $a^2+b+c=\frac{1}{a}$, multiplying by $a$ we get that: $a(a^2+b+c)=1$ and since we have $a+b+c=1$ we get $a(a^2+1-a)=1$ $a^3+a-a^2-1=0$ $(a^2+1)(a-1)=0$ so we have $a=1$ or $a=-1$, similarly for $b$ we have $(b^2+1)(b-1)=0$ since $a\neq b$ we need to have (if $a=1,b=-1$or $if a=-1,b=1$) but when we check the $2$ cases we see that $a=c$ or $b=c$ respectively,which is contradiction to our assumption.

@above I think you made a mistake, Subtracting $a^3+ab+ac=1$ by $b^3+bc+ab=1\implies a^3+ab+ac-b^3-bc-ab=0 \implies a^3+b^3=bc-ac=c(b-a)$. Where did you get that $a^3-b^3=c(a-b)$?

i mean $a^3-b^3=c(b-a)$ sorry my bad. I believe now its fine.

Multiply all the equations and set $a+b+c=s$ to get that: $$a^{3}+ab+ac-1=0, b^{3}+ba+bc-1=0, c^{3}+ca+cb-1=0 \iff a^{3}-a^{2}+sa-1=0, b^{3}-b^{2}+sb-1=0, c^{3}-c^{2}+sc-1=0$$Now if we imagine that $a\not= b\not= c$ and set $P(x)=x^{3}-x^{2}+sx-1$ then $a,b,c$ are the three roots of $P(x)$ and as the leading coefficient of $P(x)$ is $1$ so: $$P(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c)x^{2}+(ab+bc+ca)x-abc=x^{3}-x^{2}+sx-1 \implies a+b+c=1 \implies s=1 \implies P(x)=x^{3}-x^{2}+x-1$$From there though $P(x)=x^{3}-x^{2}+x-1=(x-1)(x^{2}+1)$ so $P(x)$ has only one root and the proof is complete.

Assume otherwise. Subtracting the pairs of equations gives $a^2+b-b^2-a=\frac1a-\frac1b$, or $(a-b)(a^2b+ab^2-ab+1)=0$, so $a^2b+ab^2=ab-1$. Similarly, $b^2c+bc^2=bc-1$ and $c^2a+ca^2=ca-1$. Subtracting the pairs of equations again gives $b^2c-ca^2+bc^2-c^2a=bc-ca$, or $c(b-a)(a+b+c-1)=0$, so $a+b+c=1$. Since $a^2+b+c=\frac1a$, $a^3-a^2+a-1=0$ or $(a-1)^2(a+1)=0$. So $\{a,b,c\}\subseteq\{-1,1\}$, a contradiction.

Similar... Assume contrary $a\neq b , b\neq c, c\neq a$ $a^2+b+c = \frac{1}{a} \implies a^3+ab+ac=1$, Similarly $b^3+bc+ba=1$ and $c^3+ca+cb=1$ Thus $a^3-b^3+ac-bc=0 \iff (a-b)(a^2+ab+b^2+c)=0 \implies a^2+ab+b^2+c=0$ Similarly $b^2+bc+c^2+a=0$ , Then $a^2+ab+b^2+c=b^2+bc+c^2+a \iff (a-b)(a+b+c-1)=0$ Hence $a+b+c=1\implies b+c=1-a$ Since $a^2+b+c = \frac{1}{a} \implies a^2+1-a=\frac{1}{a} \implies a^3-a^2+a-1=0 \implies (a-1)(a^2+1)=0$ Since $a^2+1 \ge 1$ we have $a-1=0$ similarly $b-1=0 , c-1=0 \implies$ Hence $a+b+c=3 \implies$ Contradiction!!! so we are done