A strange equation VI

Let $a=x+\frac {2}{x}$

Note that since $x$ is not 0, $a$ is defined for all roots. Also, the equation is equivalent to the following: $a^{2}-4a-20=0$ . To see this we divide the original equation by $x^{2}$ and add $4$ on both sides to get \[ \begin{array}{} & x^{2}-4x-16+4-\frac {8}{x}+\frac {4}{x^{2}} \\ & =x^{2}+4+\frac {4}{x^{2}}-4 (x+\frac {2}{x})-16 \\ & =a^{2}-4a-16 \\ & =4. \end{array}\] The quadratic equation $a^{2}-4a-20=0$ follows.

Now we solve it using the quadratic formula to get $a=2+2\sqrt {6}, 2-2\sqrt {6}$ . From $a=x+\frac {2}{x}$ we get two quadratic equations in the form $x^{2}-ax+2=0$ . Solving them, we have $x=1+\sqrt {6} \pm (\sqrt {2}+\sqrt {3}), 1-\sqrt {6} \pm (\sqrt {2}-\sqrt {3})$ . Only $1+\sqrt {6}+\sqrt {2}+\sqrt{3}$ is in the required form. Hence the answer is $1+2×3×6=37$

We can take x-a=√ b+√ c+√d then square it both side.then we get b, c,d as root free. After that takingb+c+d on the rootless side. Squaring again we get the eqn given in the question basically. Now equalising the coefficients of x, x^2,x^3 we get a=1 √bcd=6. Hence a+BCD=37