Solving the roots of Redlich Kwong Equation

I put a second answer to show you how the work can be done in a simpler manner.

Start with the equation of state $$P=\frac{R T}{V-b}-\frac{a}{\sqrt{T}\, V (b+V)}\tag 1$$ and develop it in terms of $V$; this leads to a cubic polynomial in $V$ $$V^3-\frac{R T}{P} V^2+ \left(\frac{a}{P \sqrt{T}}-b^2-\frac{b R T}{P}\right)V-\frac{a b}{P \sqrt{T}}=0$$ Now, at the critical point $(T=T_c, P=P_c)$, this must be identical to the development of $$(V-V_c)^3=V^3-3V_c V^2+3V_c^2V-V_c^3$$ Now, by identification $$3V_c=\frac{R T_c}{P_c}\tag 2$$ $$3V_c^2=\frac{a}{P_c \sqrt{T_c}}-b^2-\frac{b R T_c}{P_c}\tag 3$$ $$V_c^3=\frac{a b}{P_c \sqrt{T_c}}\tag 4$$ So, we can only use equations $(3)$ and $(4)$ to identify $a$ and $b$.

From $(4)$ we can express $a$ as a function of $b$ and plug it in $(3)$; this gives now a cubic equation in $b$; solve it and go back to the definition we made of $a$ as a function of $b$.

Edit

More details. From $(2)$ we have $V_c=\frac{R T_c}{3 P_c}$ this makes the cubic in $b$ to be $$b^3+\frac{b^2 R T_c}{P_c}+\frac{b R^2 T_c^2}{3 P_c^2}-\frac{R^3 T_c^3}{27 P_c^3}=0$$ Define now $b=\frac{B RT_c}{P_c}$; this makes the equation $$27 B^3+27B^2+9B-1=0$$ Using Cardano method, there is only one real root which is $$B=\frac{1}{3} \left(\sqrt[3]{2}-1\right)\implies b=\frac{1}{3} \left(\sqrt[3]{2}-1\right)\frac{ RT_c}{P_c}$$ Now, $(4)$ becomes $$\Big(\frac{R T_c}{3 P_c}\Big)^3=\frac{a b}{P_c \sqrt{T_c}}=\frac{a }{P_c \sqrt{T_c}}\frac{1}{3} \left(\sqrt[3]{2}-1\right)\frac{ RT_c}{P_c} \implies a=\frac{R^2 T_c^{5/2}}{9 \left(\sqrt[3]{2}-1\right) P_c}$$