An interesting Binomial sums ends with a quartic equation.

Before solving the given series we set $F(x)=\sum_{n\geq 0} {2n+1\choose n}x^n,\; |x|<\frac{1}{4}$ and the generating function for $F(x)$ is given by $\begin{aligned} F(x)\ = \sum_{n\geq 0}{ 2n+1\choose n}x^n\ =\frac{2}{\sqrt{(1-4x)(1+\sqrt{1-4x})}}=\frac{1}{2x\sqrt{1-4x}}-\frac{1}{2 x}\cdots(1)\end{aligned}$ now set $x=-\frac{1}{8}$ in equation 1 gives us $F(-8^{-1}$ $\sum_{n\geq 0}{2n+1\choose n} \left(-\frac{1}{8}\right)^n=4-4\sqrt{\frac{2}{3}}$ Now further differentiating equation 1 with respect to $x$ and then multiply by $x$ $xF'(x)=\frac{\sqrt{1-4x} D[x^{-1}]-D[\sqrt{1-4x} ]x^{-1}}{1-4x} +\frac{1}{2x^2}\\=\frac{1}{2x}+\frac{1}{\sqrt{(1-4x)^3}}-\frac{1}{2x\sqrt{1-4x}}$ now setting $x=\frac{1}{8}$ in equation 1 gives us $F'(8^{-1})=\sum_{n\geq 0}\frac{n}{8^n}{2n+1\choose n}=\sum_{n\geq 0}\frac{n}{2^{3n}}{ 2n+1\choose n}\\=4+2\sqrt 2-4\sqrt 2=4-2\sqrt2$ and hence $\beta = \frac{\sqrt 2 xF'(8^{-1} )+F(-8^{-1})}{\sqrt 2 xF'(8^{-1} )-F(-8^{-1})}=\frac{\sqrt2(4-2\sqrt 2)+4-4\sqrt{\frac{2}{3}}}{\sqrt2(4-2\sqrt 2)-4+4\sqrt{\frac{2}{3}}}=\frac{1}{2}\left(1+\sqrt{3}+\sqrt{6}\right)$ Now we will show that $B=2\beta^4-4\beta^3-6\beta^2+8\beta=1$ Now using multinomial theorem we can deduce that $\begin{aligned}2\beta^4\ = 26+18\sqrt 2 + 11\sqrt 3 + 8\sqrt 6 \\ 4\beta^3 =14+9\sqrt 2 +12\sqrt 3 +9\sqrt 6\\ 6\beta^2=15+9\sqrt 2 +3\sqrt 3 +3\sqrt 6\\ 8\beta = 4+4\sqrt 3+4\sqrt 6\end{aligned}$ and hence $B= 26-15-14+4=30-29=1$