A strange equation II

Let's substitute $x=y+1$ in order to delete the $x^{2}$ coefficient: $(y+1)^{3}-3(y+1)^{2}+1=0$ $y^{3}-3y-1=0$

Now, there comes the tricky part. Let's assume the identity $(u+v)^{3}-3uv(u+v)-(u^{3}+v^{3})=0$ and make it match with the equation: $y=u+v$ $-3uv=-3 \Leftrightarrow uv=1$ $-(u^{3}+v^{3})=-1 \Leftrightarrow u^{3}+v^{3}=1$

Now, another trick: cube both sides of the first equation: $uv=1 \Leftrightarrow (u)^{3}(v)^{3}=1$

With this, we have an equation system that matches with a second degree equation, so let's $z_{1}=u^{3}$ and $z_{2}=v^{3}$ : $z^{2}-z+1=0$

Use the quadratic formula: $z_{1,2}=\dfrac{1\pm i\sqrt{3}}{2}$

And substitute back for $u$ and $v$ : $u=\sqrt[3]{\dfrac{1+i\sqrt{3}}{2}}$ $v=\sqrt[3]{\dfrac{1-i\sqrt{3}}{2}}$ Where $i=\sqrt{-1}$ is the imaginary unit.

Express them in the polar form: $u=\sqrt[3]{e^{i\left(\dfrac{\pi}{3}+2\pi k \right)}} \Leftrightarrow u=e^{\dfrac{\pi i}{9} \left(6k+1\right)}$ $v=\sqrt[3]{e^{-i\left(\dfrac{\pi}{3}+2\pi k \right)}} \Leftrightarrow v=e^{-\dfrac{\pi i}{9} \left(6k+1\right)}$ Where $k={0,1,2}$

Reverse to the rectangular form and convert from radians to degrees: $u=\cos(20°(6k+1))+i \sin(20°(6k+1))$ $v=\cos(20°(6k+1))-i \sin(20°(6k+1))$

Now, for $k=0$ : $u=\cos(20°)+i \sin(20°)$ $v=\cos(20°)-i \sin(20°)$ So, $y=u+v$ : $y=\cos(20°)+i \sin(20°)+\cos(20°)-i \sin(20°)$ $y=2\cos(20°)$ And, $x=y+1$ : $\boxed{x=2\cos(20°)+1}$

Now, for $k=1$ : $u=\cos(140°)+i \sin(140°) \leftrightarrow u=-\cos(40°)+i \sin(40°)$ $v=\cos(140°)-i \sin(140°) \leftrightarrow v=-\cos(40°)-i \sin(40°)$ So, $y=u+v$ : $y=-\cos(40°)+i \sin(40°)-\cos(40°)-i \sin(40°)$ $y=-2\cos(40°)$ And, $x=y+1$ : $\boxed{x=-2\cos(40°)+1}$

And finally, for $k=2$ : $u=\cos(260°)+i \sin(260°) \leftrightarrow u=-\cos(80°)-i \sin(80°)$ $v=\cos(260°)-i \sin(260°) \leftrightarrow v=-\cos(80°)+i \sin(80°)$ So, $y=u+v$ : $y=-\cos(80°)-i \sin(80°)-\cos(80°)+i \sin(80°)$ $y=-2\cos(80°)$ And, $x=y+1$ : $\boxed{x=-2\cos(80°)+1}$

We obtained three solutions for $x$ , and after ordering them, we have: $x_{1}=-2\cos(40°)+1$ $x_{2}=-2\cos(80°)+1$ $x_{3}=2\cos(20°)+1$

So, $a=1$ , $b=c=d=2$ , $e=40$ , $f=80$ and $g=20$ , and $a+b+c+d+e+f+g=\boxed{147}$