A strange equation IV

First, we delete the $x^{4}$ coefficient by making the substitution $x=y+3$ : $(y+3)^{5}-15(y+3)^{4}+70(y+3)^{3}-90(y+3)^{2}-55(y+3)-32\sqrt{2}+57=0$ $y^{5}-20y^{3}+80y-32\sqrt{2}=0$ Note that also the $x^{2}$ coefficient was deleted.

Now, pay attention to the following identity: $(u+v)^{5}-5uv(u+v)^{3}+5(uv)^{2}(u+v)-(u^{5}+v^{5})=0$ and observe that it matches with the equation. That special case of equation is known as de Moivre's quintic , which is solvable over the radicals. Now, compare the coefficients of the equation in $y$ with the identity, to obtain an equation system: $y=u+v$ $-5uv=-20 \leftrightarrow uv=4$ $-(u^{5}+v^{5})=-32\sqrt{2} \leftrightarrow u^{5}+v^{5} = 32\sqrt{2}$ Raise both sides of the second equation to the fifth power: $(u)^{5}(v)^{5}=1024$

And then, check that we have, by Vieta's formulas, a second degree equation with roots $u^{5}$ and $v^{5}$ . So, let $z_{1}=u^{5}$ and $z_{2}=v^{5}$ : $z^{2}-32\sqrt{2}z+1024=0$ Solve it by the quadratic formula to obtain the two roots: $z_{1,2}=16\sqrt{2}\pm16i\sqrt{2}$ And express them in polar form: $z_{1,2}=32e^{\pm(\dfrac{\pi i}{4}+2\pi ki)}$ $z_{1,2}=32e^{\pm \dfrac{\pi i(8k+1)}{4}}$ Solve back for $u$ and $v$ : $u,v=\sqrt[5]{32e^{\pm \dfrac{\pi i(8k+1)}{4}}}$ Simplify and use exponent laws: $u,v=2e^{\pm \dfrac{\pi i(8k+1)}{20}}$ Reverse to the rectangular form and express the argument in degrees: $u,v=2[\cos(9°(8k+1)) \pm i \sin(9°(8k+1))]$

Now, remember that $y=u+v$ , so: $y=4\cos(9°(8k+1))$ Note that the sine part cancells because $u$ and $v$ are complex conjugates.

Also, remember that we have $k=0,1,2,3,4$ , because we have five fifth roots. So, let's compute for every value of $k$ , applying some trigonometric identities to express every argument in the range $[0°,90°]$ :

For $k=0$ : $y=\boxed{4\cos(9°)}$

For $k=1$ : $y=\boxed{4\cos(81°)}$

For $k=2$ : $y=\boxed{-4\cos(27°)}$

For $k=3$ : $y=-4\cos(45°)=\boxed{-2\sqrt{2}}$

For $k=4$ : $y=\boxed{4\cos(63°)}$

Finally, remember that $x=y+3$ , so we obtain the solutions to the original equation, and after ordering them in ascending form:

$\boxed{x_{1}=3-4\cos(27°)}$

$\boxed{x_{2}=3-2\sqrt{2}}$

$\boxed{x_{3}=3+4\cos(81°)}$

$\boxed{x_{4}=3+4\cos(63°)}$

$\boxed{x_{5}=3+4\cos(9°)}$

Hence, $a=3$ , $b=4$ , $c=2$ , $d=27$ , $e=81$ , $f=63$ , $g=9$ and $d+e+f+g-a-c-b=\boxed{171}$ .

The first step should be to remove the $x^{ 4 }$ term by substituting $x= t + 3$ The corresponding equation should be $t^{ 5 }-20t^{ 3 }+80t-32√2 = 0$ As $cos(5x)=16cos^{ 5 }(x)-20cos^{ 3 }(x)+5cos(x)$ the equation can be reduced to $cos(5u)= 1/\sqrt { 2 }$ by substituting $t$ as $4cos(u)$ The solutions will be $3 + 4cos(9), 3 + 4cos(81), 3 - 4cos(27), 3 - 4cos(45), 3 + 4cos(63)$