Solving the roots of Redlich Kwong Equation Good evening everyone.
After studying some equations of state, I've read about the mathematical steps formulated to model some. In the particular case of Van der Waals, where
$$P=\frac{RT}{v-b}-\frac{a}{v^2}$$
For those not familiar with this expression, this equation describes general characteristics of a real gas, P standing for pressure, R for the universal gas constant, T for temperature, v for molar volume (Yep, I know the little line is missing, sorry!), and a and b are characteristic constants for the gas . 
There are some special values for Temperature, Pressure and Molar volume. Finding them, apparently, helps me in my quest for a and b. I am told the following steps to do so in VdW equation:
1) Derive the function twice.
2) Solve RT for each derivative and make them equal to zero.Make an equation of of the derivative and the second derivative, then solve for v
3) Plug the value of V in the first derivative, and solve for T, now called critical temperature. 3
4) Plug the values of V and T in the original function, thus obtaining the critical pressure. 4
5) Solve "a" for critical T and critical P, then making the equation and obtaining "b"[5]
6) Plug "b" in critical T, now obtaining "a"[6]
Now, back to my problem, I'm looking for "a" and "b" again, but now in the Redlich Kwong equation:
$$P=\frac{RT}{v-b}-\frac{a}{\sqrt(T)*(v^2+vb)}$$
I've already derived the function twice and solved for RT:
$$P'=\frac{-RT}{(v-b)^2}+\frac{a(2v+b)}{\sqrt(T)*((v^2+vb)^-2)}=0$$
$$RT=\frac{a(2v+b)*((v-b)^2}{\sqrt(T)*((v^2+vb)^-2)}$$
$$P''=\frac{2RT}{(v-b)^-3}+\frac{2a}{\sqrt(T)*((v^2+vb)^-2)}-\frac{2a*(2v+b)^2}{\sqrt(T)*((v^2+vb)^-3)}=0$$
$$RT=\frac{a}{\sqrt(T)}*((v-b)^3)*(\frac{(2v+b)^2}{((v^2)+vb)^3}-\frac{1}{(v^2)+vb)^2})$$
After putting the two =0, I've obtained this far:
$$3v+2b=(v-b)*((2v+b)^2)((v^2)+vb)$$
As it isn't as simple as in the VdW case, I certainly don't know where to go now in order to obtain a value of v based on b. I would be so grateful if someone checked this, thank you! 
 A: 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 $$\Big(\frac{R T_c}{3 P_c}\Big)^3=\frac{a b}{P_c
   \sqrt{T_c}}=\frac{a }{P_c
   \sqrt{T_c}}\frac{1}{3} \left(\sqrt[3]{2}-1\right)\frac{ RT_c}{P_c} \implies a=\frac{R^2 T_c^{5/2}}{9 \left(\sqrt[3]{2}-1\right) P_c}$$
A: Even if there exist much simpler procedures to get the characteristic constants of cubic equations of state, let us do it using almost what you have been suggested.
Using $$P=\frac{R T}{V-b}-\frac{a}{\sqrt{T} V (b+V)}\tag 1$$ $$P'=\frac 1{\sqrt T}\Big(\frac{a (b+2 V)}{V^2 (b+V)^2}-\frac{R T^{3/2}}{(b-V)^2}\Big)\tag2$$ $$P''=\frac 1{\sqrt T}\Big(\frac{a \left(\frac{1}{(b+V)^3}-\frac{1}{V^3}\right)}{b}+\frac{R T^{3/2}}{(V-b)^3}\Big)\tag3$$ SInce, at the critical point, $P'=P''=0$, equations $(2)$ and $(3)$ can rewrite as $$\frac{R T^{3/2}}{(V-b)^2}=\frac{a (b+2 V)}{V^2 (b+V)^2}\tag 4$$ $$\frac{R T^{3/2}}{(V-b)^3}=-\frac{a \left(\frac{1}{(b+V)^3}-\frac{1}{V^3}\right)}{b}\tag 5$$ Now, making the ratio ($(4)$ divided by $(5)$) and simplifying, we then get $$V-b=\frac{V (b+V) (b+2 V)}{b^2+3 b V+3 V^2}\tag 6$$ Cross multiplying and   simplifying, we then get $$-b^3-3 b^2 V-3 b V^2+V^3=0$$ which is an homogeneous polynomial of degree $3$. Setting $V=x b$, we then need to solve $$x^3-3 x^2-3 x-1=0$$ Using Cardano method, the only real solution is given by $$x=1+2^{2/3}+2^{1/3}$$
I am sure that you can continue from here.
