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 $c \neq 0$, 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.

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 $c \neq 0$, 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.