A strange equation III

We first depress the cubic equation.

$\begin{aligned} x^3 + 3x^2 - 51x + 54\sqrt3 - 107 & = & (x+1)^3 - 54x - 108 + 54 \sqrt3 \\ & = & (x + 1)^3 - 54(x + 1) + (54 \sqrt3 - 54), \text{ let } y=x+1 \\ & = & y^3 - 54y + (54 \sqrt3 - 54) \\ \end{aligned}$

We want to solve the equation via trigonometric equations. Recall the triple angle formula $\cos (3 \Theta) = 4 \cos^3 ( \Theta) - 3 \cos (\Theta )$ , so we must have another substitution with $A>0$ , $y = A \cos (\Theta )$ such that the ratio coefficient of $\cos^3 (\Theta)$ to $\cos (\Theta)$ for $y^3$ and $54y$ is $4:3$ after substitution.

$\begin{aligned} y^3 : 54y & = & 4 : 3 \\ A^3 : 54A & = & 4 : 3 \\ \frac {A^3}{54A} & = & \frac {4}{3} \\ A & = & \sqrt{72} \\ \end{aligned}$

Thus the substitution $y = \sqrt{72} \cos ( \Theta )$

$\begin{aligned} \sqrt{72} \cdot (72 \cos^3 (\Theta) - 54 \cos (\Theta) ) +(54 \sqrt3 - 54) & = & 0 \\ \sqrt{72} \cdot 18 \cdot (4 \cos^3 (\Theta) - 3 \cos (\Theta) ) & = & 54 - 54 \sqrt 3 \\ \sqrt{72} \cdot \cos (3 \Theta) & = & (3 - 3 \sqrt 3) \\ \cos (3 \Theta) & = & \frac {3 - 3 \sqrt 3 }{\sqrt {72}} \\ & = & - \frac {\sqrt{3} - 1}{2 \sqrt2 } \\ \end{aligned}$

With the double angle formula $\cos ( 2 \phi ) = 1 - 2 \sin^2 ( \phi )$ ,

$\begin{aligned} 1 - 2 \sin^2 ( 15^\circ) & = & \cos (30^\circ ) \\ 1 - 2 \sin^2 ( 15^\circ) & = & \frac {\sqrt 3}{2} \\ \sin ( 15^\circ) & = & \sqrt{ \frac {1 - \frac {\sqrt 3}{2} }{2} } \\ \cos ( 75^\circ) & = & \frac {\sqrt{4 - 2 \sqrt3 }}{2 \sqrt2} \\ & = & \frac {\sqrt{1^2 + (\sqrt 3)^2 - 2 \cdot 1 \cdot \sqrt3 }}{2 \sqrt2} \\ & = & \frac {\sqrt3 - 1}{2 \sqrt2} \\ \cos ( 105^\circ) & = & - \frac {\sqrt3 - 1}{2 \sqrt2} \\ \cos ( 105^\circ), \cos ( 360^\circ - 105^\circ), \cos ( 360^\circ + 105^\circ) & = & - \frac {\sqrt3 - 1}{2 \sqrt2} \\ \cos ( 105^\circ), \cos ( 255^\circ), \cos ( 465^\circ) & = & - \frac {\sqrt3 - 1}{2 \sqrt2} \\ \end{aligned}$

So the $3$ smallest positive values of $3 \Theta$ is $105^\circ, 255^\circ, 465^\circ$ , equivalently $\Theta = 35^\circ, 85^\circ, 155^\circ$

Because we want the angles to be positive acute angles, we change the last angle to $180^\circ - 155^\circ = 25^\circ$

$\begin{aligned} y & = & \sqrt{72} \cos (35^\circ), \sqrt{72} \cos (85^\circ), -\sqrt{72} \cos (25^\circ) \\ & = & 6\sqrt2 \cos (35^\circ), 6\sqrt2 \cos (85^\circ), -6\sqrt2 \cos (25^\circ) \\ x + 1 & = & 6\sqrt2 \cos (35^\circ), 6\sqrt2 \cos (85^\circ), -6\sqrt2 \cos (25^\circ) \\ x & = & 6\sqrt2 \cos (35^\circ) - 1, 6\sqrt2 \cos (85^\circ) - 1, -6\sqrt2 \cos (25^\circ) - 1\\ \end{aligned}$

Hence, $a = 6, b = 2, c = 1, \alpha = 25, \beta = 35, \theta= 85 \Rightarrow \frac { \alpha + \beta + \theta - 1}{a+b+c} = \boxed{16}$

Let's make the substitution $x=y-1$ to delete the $x^{2}$ coefficient: $(y-1)^{3}+3(y-1)^{2}-51(y-1)+54\sqrt{3}-107=0$ $y^{3}-54y+54\sqrt{3}-54=0$

Now, consider the identity $(u+v)^{3}-3uv(u+v)-(u^{3}+v^{3})=0$ and make it coincide with the equation: $y=u+v$ $-3uv=-54 \leftrightarrow uv=18$ $-(u^{3}+v^{3})=54\sqrt{3}-54 \leftrightarrow u^{3}+v^{3}=54-54\sqrt{3}$

And cube both sides of the second equation: $uv=18 \leftrightarrow (u)^{3}(v)^{3}=5832$

Observe that the second and the third equation matches, by Vieta's formula, with a second-degree equation with roots $u^{3}$ and $v^{3}$ . So, let $z_{1}=u^{3}$ and $z_{2}=v^{3}$ : $z^{2}+(54\sqrt{3}-54)z+5832=0$ .

Use the quadratic formula to solve for $z_{1,2}$ : $z_{1,2}=27-27\sqrt{3}\pm i(27\sqrt{3}+27)$ Where $i=\sqrt{-1}$ is the imaginary unit.

Now, let's express both values of $z$ in polar form, but in this case the modulus $R$ and the argument $\delta$ aren't very evident, so let's compute them: $R=\sqrt{(27-27\sqrt{3})^{2}+(27\sqrt{3}-27)^{2}}$ $R=54\sqrt{2}$ $\tan \delta=\dfrac{27\sqrt{3}+27}{27-27\sqrt{3}}$ $\tan \delta=-(2+\sqrt{3})$ We're lucky because we can compute exactly $\delta$ : $\delta=\dfrac{7\pi}{12}$

Now, express $z_{1,2}$ in polar form: $z_{1,2}=Re^{i(\delta+2\pi k)}$ $z_{1,2}=54\sqrt{2}e^{\pm\dfrac{\pi i}{12}(7+24k)}$

And substitute back for $u$ and $v$ : $u=\sqrt[3]{54\sqrt{2}e^{\dfrac{\pi i}{12}(7+24k)}}$ $u=3\sqrt{2}e^{\dfrac{\pi i}{36}(7+24k)}$ $v=\sqrt[3]{54\sqrt{2}e^{-\dfrac{\pi i}{12}(7+24k)}}$ $v=3\sqrt{2}e^{-\dfrac{\pi i}{36}(7+24k)}$ And express now in rectangular form and convert radians to degrees: $u=3\sqrt{2}[\cos(5°(7+24k))+i \sin(5°(7+24k))]$ $v=3\sqrt{2}[\cos(5°(7+24k))-i \sin(5°(7+24k))]$ For $k={0,1,2}$ .

Now, remember that $y=u+v$ and $x=y-1$ , so let's compute for the three values of $k$ .

For $k=0$ : $u=3\sqrt{2}[\cos(35°)+i \sin(35°)]$ $v=3\sqrt{2}[\cos(35°)-i \sin(35°)]$ $y=3\sqrt{2}[\cos(35°)+i \sin(35°)]+3\sqrt{2}[\cos(35°)-i \sin(35°)]$ $y=6\sqrt{2}\cos(35°)$ $\boxed{x=6\sqrt{2}\cos(35°)-1}$

For $k=1$ : $u=3\sqrt{2}[\cos(155°)+i \sin(155°)] \leftrightarrow u=3\sqrt{2}[-\cos(25°)+i \sin(25°)]$ $v=3\sqrt{2}[\cos(155°)-i \sin(155°)] \leftrightarrow u=3\sqrt{2}[-\cos(25°)-i \sin(25°)]$ $y=3\sqrt{2}[-\cos(25°)+i \sin(25°)]+3\sqrt{2}[-\cos(25°)-i \sin(25°)]$ $y=-6\sqrt{2}\cos(25°)$ $\boxed{x=-6\sqrt{2}\cos(25°)-1}$

For $k=2$ : $u=3\sqrt{2}[\cos(275°)+i \sin(275°)] \leftrightarrow u=3\sqrt{2}[\cos(85°)-i \sin(85°)]$ $v=3\sqrt{2}[\cos(275°)-i \sin(275°)] \leftrightarrow u=3\sqrt{2}[\cos(85°)+i \sin(85°)]$ $y=3\sqrt{2}[\cos(85°)-i \sin(85°)]+3\sqrt{2}[\cos(85°)+i \sin(85°)]$ $y=6\sqrt{2}\cos(85°)$ $\boxed{x=6\sqrt{2}\cos(85°)-1}$

Now, order the three solutions for $x$ : $x_{1}=-6\sqrt{2}\cos(25°)-1$ $x_{2}=6\sqrt{2}\cos(85°)-1$ $x_{3}=6\sqrt{2}\cos(35°)-1$

So, $\alpha=25$ , $\beta=85$ , $\theta=35$ , $a=6$ , $b=2$ , $c=1$ and $\dfrac{\alpha+\beta+\theta-1}{a+b+c}=\dfrac{144}{9}=\boxed{16}$ .

With f(x) = x^3 + 3 x^2 - 51 x + 54 sqrt(3) - 107,

Let x = y + k,

f(k) = k^3 + 3 k^2 - 51 k + 54 sqrt(3) - 107

f'(x)/ 1! = 3 k^2 + 6 k - 51

f''(x)/ 2! = (6 k + 6)/ 2 = 3 k + 3

f'''(x)/ 3! = 6/ 6 = 1

With f''(x)/ 2! = 0, k = -1,

y^3 - 54 y + 54 [sqrt(3) - 1] = 0 which can apply cubic formula. f^n (x)/ n! can minimize mistake and able to allow computer to do.

Applying sqrt[4 - 2 sqrt(3)] = sqrt(3) - 1,

x = -3 {cbrt[sqrt(3) - 1 - j (sqrt(3) + 1)] + cbrt[sqrt(3) - 1 + j (sqrt(3) + 1)] - 1;

|sqrt(3) - 1 - j (sqrt(3) + 1)| = |sqrt(3) - 1 + j (sqrt(3) + 1)| = 2 sqrt(2);

Arg[sqrt(3) - 1 - j (sqrt(3) + 1)] = - 75 deg while Arg[sqrt(3) - 1 + j (sqrt(3) + 1)] = 75 deg;

Finally,

x1 = - 6 sqrt(2) Cos 25 deg - 1

x2 = 6 sqrt(2) Cos 85 deg - 1

x3 = 6 sqrt(2) Cos 35 deg - 1

a = 6, b = 2 and c = 1.

(A + B + T - 1)/ (a + b + c) = (25 + 85 + 35 - 1)/ (6 + 2 + 1) = 144/ 9 = 16