A strange equation V

First, substitute $x=y-4$ to delete some coefficients. We obtain: $(y-4)^7+28(y-4)^6+308(y-4)^5+1680(y-4)^4+4704(y-4)^3+6272(y-4)^2+3136(y-4)-128\sqrt{2}+256=0$ $y^7-28y^5+224y^3-448y-128\sqrt{2}=0$ We deleted the $x^6$ , $x^4$ and $x^2$ coefficients. But now, how to solve it?

This identity will help us: $(u+v)^7-7uv(u+v)^5+14(uv)^2(u+v)^3-7(uv)^3(u+v)-(u^7+v^7)=0$ . But, where did it come from? Well, first observe that it matches perfectly with our equation in $y$ , and this is a special case of seven degree equation, named De Moivre's septic . So, now, form an equation system by comparing the coefficients: $y=u+v$ $-7uv=-28 \leftrightarrow uv=4$ $-(u^7+v^7)=-128\sqrt{2} \leftrightarrow u^7+v^7=128\sqrt{2}$ Raise both sides of the second equation to the $7^{th}$ power: $(u)^7(v)^7=16384$ $u^7+v^7=128\sqrt{2}$ These two equations form a second degree equation by Vieta's formulas, with roots $u^7$ and $v^7$ . So, let $z_{1}=u^7$ and $z_{2}=v^{7}$ : $z^2-128\sqrt{2}+16384=0$ Use the quadratic formula and simplify: $z_{1,2}=64\sqrt{2}\pm 64i\sqrt{2}$ Transform that expression into the polar form: $z_{1,2}=128e^{\pm\dfrac{\pi i(8k+1)}{4}}$ Now, obtain $u$ and $v$ : $u=\sqrt[7]{128e^{\dfrac{\pi i(8k+1)}{4}}}=2e^{\dfrac{\pi i(8k+1)}{28}}$ $v=\sqrt[7]{128e^{-\dfrac{\pi i(8k+1)}{4}}}=2e^{-\dfrac{\pi i(8k+1)}{28}}$ And reverse to the rectangular form, still in radians: $u=2\left(\cos\left(\dfrac{\pi(8k+1)}{28}\right)+i \sin\left(\dfrac{\pi(8k+1)}{28}\right)\right)$ $v=2\left(\cos\left(\dfrac{\pi(8k+1)}{28}\right)-i \sin\left(\dfrac{\pi(8k+1)}{28}\right)\right)$

As, we stated at the beginning, $y=u+v$ , so: $y=4\cos\left(\dfrac{\pi(8k+1)}{28}\right)$ (Note that the imaginary part cancelled, because $u$ and $v$ are complex conjugates.) And, $x=y-4$ , so: $\boxed{x=-4+4\cos\left(\dfrac{\pi(8k+1)}{28}\right)}$

And finally, compute the seven solutions for $k=0,1,2,3,4,5,6$ and order them. I will use some trigonometric identities to always express the angle in the range $[0,\frac{\pi}{2}]$ : $\boxed{x_{1}=-4-4\cos(\frac{3\pi}{28})}$ $\boxed{x_{2}=-4-4\cos(\frac{5\pi}{28})}$ $\boxed{x_{3}=-4-4\cos(\frac{11\pi}{28})}$ $\boxed{x_{4}=-4-4\cos(\frac{13\pi}{28})}$ $\boxed{x_{5}=-4+4\cos(\frac{9\pi}{28})}$ $x_{6}=-4+4\cos(\frac{\pi}{4}) = \boxed{-4+2\sqrt{2}}$ $\boxed{x_{7}=-4+4\cos(\frac{\pi}{28})}$

Hence, $a=4$ , $b=4$ , $c=3$ , $d=28$ , $e=5$ , $f=11$ , $g=13$ , $h=9$ , $j=2$ , $k=1$ and $a+b+c+d+e+f+g+h+j+k=\boxed{80}$

x^7 + p x^5 + (2/ 7) p^2 x^3 + (1/ 49) p^3 x + q = 0

x^6 + p x^4 + (1/ 4) p^2 x^2 + q = 0 {Forgot when I added into my memo.}

x^5 + p x^3 + (1/ 5) p^2 x + q = 0

x^3 + p x + q = 0

These are outcome of binomial expansions. I am thinking of introducing Cube_Root (p, q) as a basic function for cubic equation mentioned above.

x^4 + p x^2 + q x + r = 0 and x^5 + p x^3 + q x^2 + (p^2/ 5) x + r = 0 are Trinomials as I named them, but there is nothing to gain. How is your conclusion for x^5 + p x^4 + q x^3 + r x^2 + s x + t = 0?