Find all $(a, b, c) \in \mathbb R^3$ satisfying the following system of equations: $$ \begin{cases} a^{2} = \dfrac{ {b}^3 + 9 \sqrt{3} }{3b} = \dfrac{ {c}^3 + 16 }{3c} \\ \\ b^{2} = \dfrac{ {a}^3 - 10 }{3a} = \dfrac{ {c}^{3} + 28 }{3c} \\ \\ c^{2} = \dfrac{ {b}^{3} + 45 \sqrt{3} }{3b} = \dfrac{ {a}^3 - 88 }{3a} \end{cases}$$
One possible way to solve this system of equations is as follows:
From the first equation, we can write: \begin{align} a^2 &= \dfrac{b^3 + 9\sqrt{3}}{3b} \\ 3a^2 b &= b^3 + 9\sqrt{3} \end{align}
Simplifying and rearranging, we get: $$b^3 - 3a^2 b + 9\sqrt{3} = 0$$
This is a cubic equation in $b$. We can solve it using any method for solving cubic equations, such as Cardano's formula or numerical methods. By solving this I get \begin{align} a & \approx 1 \\ b & \approx \sqrt[3]{-6\sqrt{3} - 3} + \sqrt[3]{-6\sqrt{3} + 7} \\ c & \approx -0.147.\end{align}
But the answer of this question is $a=-2 , b=\sqrt 3, c=-4$. What are the other possible ways to solve this question?