Open Contests

From the third equation we get $b=-\frac{1}{a}$. By substituting this in the first and second equations we obtain a new system: $a^3-\frac{1}{a}=4c$, $a-\frac{1}{a^3}=c$.

If $c=0$, we have $a^3=\frac{1}{a}$ and $a^4=1$, whence $a=1$ or $a=-1$, since $a=0$ is not possible. We have respectively $b=-1$ and $b=1$.

If $c\ne 0$, then by dividing in this system the sides of the first equation by the respective sides of the second equation, we obtain $(a^3-\frac{1}{a}):(a-\frac{1}{a^3})=4$. As $a^3-\frac{1}{a}=a^2\cdot (a-\frac{1}{a})$, this is equivalent to $a^2=4$, whence $a=\pm 2$. If $a=2$, then $b=-\frac{1}{2}$ and $c=\frac{15}{8}$. The case $a=-2$ gives the same solution with opposite signs.

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Alternative solution.

By adding the first and second equation we get $a(a^2+1)+b(b^2+1)=5c$. Since $1=-ab$ by the third equation, this equation is equivalent to $a(a^2-ab)+b(b^2-ab)=5c$ or, equivalently, $(a^2-b^2)(a-b)=5c$.

By subtracting the second equation from the first equation in the initial system we obtain $a(a^2-1)-b(b^2-1)=3c$ and by similarly substituting from the third equation we obtain $a(a^2+ab)-b(b^2+ab)=3c$ or, equivalently, $(a^2-b^2)(a+b)=3c$.

Hence if $c=0$, then $a-b=0$ or $a+b=0$. By the third equation $a=b$ is not possible. The case $a=-b$ gives two possibilities $a=1,b=-1$ and $a=-1,b=1$.

If $c\ne 0$, we obtain $\frac{a+b}{a-b}=\frac{3}{5}$, whence $a=-4b$. By substituting into the third equation of the initial system we obtain $-4b^2=-1$, whence $b=\pm \frac{1}{2}$. If $b=\frac{1}{2}$, then $a=-2$ and $c=-\frac{15}{8}$. The case $b=-\frac{1}{2}$ gives the same solution with opposite signs.