Since nobody has posted a solution yet, I will give my bashy solution (which, I think, is hard but possible to give in the exam without computational aid).

danepale wrote: Find all triples $(a, b, c)$ of real numbers such that $$ a^2 + ab + c = 0, $$$$b^2 + bc + a = 0, $$$$c^2 + ca + b = 0.$$ Sum all the equalities then getting $\frac{1}{2}\sum_{cyc}(a+b)^2 +a+b+c=0$ So it is apparent that at least $\sum a\leq0$ as if it doesn't achieve, then apparently $LHS>0$, a contradiction and WLOG, let $a \leq b \leq c$(let it as a WLOG as $a \leq c \leq b$ could make some distraction but neglect in this case) now, firstly prove that $c\leq 0$ $c=-a(a+b)$ If $c>0$, $a+b>0$ as $a<0$ is apparent by the condition given but this is contradiction to the original property $a+b+c<0$(so double contradiction of $a+b<0$ and $a+b+c<0$ by letting $c>0$), so $c \leq 0$ case 1.c=0 b=0 so a=0 then over 2.c<0 $a^2+ab+c=0$ isn't possible to be achieved if $a \leq b <c$(so, in case the equality doesn't achieve) by the absolute value between $a^2+ab$ and $c$, which the left one is apparently bigger so let $a \leq b=c$, then $a=-2c^2$(by the 2nd equation) and $-2c^3+c^2+c=0$ by the 3rd equation so $2c^2-c-1=0$ and the answers are $c=1$ and $c= -\frac{1}{2}$ but by the property of $c<0$, $c=-\frac{1}{2}$ and so does $a$. So the answers are $a=b=c=0$, and $a=b=c=-\frac{1}{2}$ quite long for the answer hmm...

danepale wrote: Find all triples $(a, b, c)$ of real numbers such that $$ a^2 + ab + c = 0, $$$$b^2 + bc + a = 0, $$$$c^2 + ca + b = 0.$$ in case of $a \leq c \leq b$, the result comes so easy as $2b^2+a \geq 0$ so $a \geq -2b^2$ and $bc-b^2 \geq 0$ so $b=c$ then a lot easier stuff it becomes.(All properties came out from the substitution to the 2nd equation) btw case of $a \leq b \leq c$ and $a \leq c \leq b$ doesn't matter to the order of the $a,b,c$...

If one of $a, b, c$ is $0$. All are $0$. This gives $(a,b,c)=(0,0,0)$. Assume $a,b,c$ are nonzero. 1. $a, b, c$ are all positive. Not happening. 2. $a$ positive, $b,c$ negative. $b^2+bc+a>0$. Not happening. 3. Two positive, one negative. WLOG $a, b> 0$ and $c<0$. This gives $b(b+c)=-a$ and $c(c+a)=-b$, so $b+c<0$ and $c+a>0$, which gives $a>b$. We also have $a+b+c=(a+b)(1-a)=(b+c)(1-b)=(c+a)(1-c) > 0$. This gives us $1-b<0$ and $1-a>0$ so $a<1<b$. Not happening. 4. Three negatives. Replace everything with $x=-a$, $y=-b$ and $z=-c$. This forces $x^2+xy=z$, $y^2+yz=x$, $z^2+zx=y$. WLOG $x=\text{min}(x,y,z)$. $y(y+z)=x$ gives $y+z \le 1$. $x(x+y)=z$ gives $x+y \ge 1$. This gives $x=z$ and $x+y=y+z=1$. Also we have $x=y$ by plugging $y+z=1$. So $x=y=z=\frac{1}{2}$. This works.