An equation with Golden Ratio

If we write $\alpha = \sqrt[3]{\tfrac{x+\varphi}{x-\varphi}}$ , then we note that $\alpha \neq 1$ and $x = \tfrac{\alpha^3 + 1}{\alpha^3 - 1}\varphi$ . Now $\begin{aligned} \alpha + \alpha^{-3} & = \; 2 \\ \alpha^4 - 2\alpha^3 + 1 &= \; 0 \\ (\alpha-1)(\alpha^3 - \alpha^2 - \alpha - 1) & = \; 0 \end{aligned}$ Thus $\alpha^3 - \alpha^2 - \alpha - 1 = 0$ , and hence $\begin{aligned} \alpha^3 & = \; \alpha^2 + \alpha + 1 \; = \; \frac{\alpha^3 - 1}{\alpha - 1} \\ \frac{2}{\alpha -1} & = \; \frac{2\alpha^3}{\alpha^3 - 1} \; = \; \frac{\alpha^3 + 1}{\alpha^3 - 1} + 1 \; = \; \frac{x + \varphi}{\varphi} \end{aligned}$ This means that $\alpha \; = \; \frac{2\varphi}{x + \varphi} + 1 \; = \; \frac{x + 3\varphi}{x + \varphi}$ and hence, since $\alpha^3 - \alpha^2 - \alpha - 1 = 0$ , $\begin{aligned} (x + 3\varphi)^3 - (x +3\varphi)^2(x + \varphi) - (x +3\varphi)(x + \varphi)^2 - (x + \varphi)^3 &= \; 0 \\ x^3 + 3\varphi x^2 -\varphi^2x - 7\varphi^3 & = \; 0 \\ (x^3 - x - 7) + (3x^2 - x - 14)\varphi & = \; 0 \end{aligned}$ and so, finally, since $\varphi^2 - \varphi - 1 = 0$ , $\begin{aligned} (x^3 - x - 7)^2 + (x^3 - x- 7)(3x^2 - x - 14) - (3x^2 -x - 14)^2 &= \; 0 \\ x^6 + 3x^5 - 12x^4 - 25x^3 + 64x^2 + 7x - 49 & = \; 0\end{aligned}$ making the answer $\boxed{7}$ .

Let $\cfrac{x+\phi}{x-\phi}=y$ .

$\sqrt[3]{y}+\frac{1}{y}=2 \\ \Longrightarrow y\sqrt[3]{y}+1=2y \\ \Longrightarrow y^{\frac{4}{3}}=2y+1 \\ \Longrightarrow y^4 = (2y+1)^3 = 8y^3-12y^2+6y-1 \\ \Longrightarrow y^4-8y^3+12y^2-6y+1=0$

There are $2$ real roots for this polynomial, those being $1$ approximately $6.2223$

We can only solve for $x$ for the second one, because $\frac{x+\phi}{x-\phi}=1$ doesn't exist ever. We must try use other root:

$\cfrac{x+\phi}{x-\phi} = 6.2223\dots$ $x+\phi = x\cdot 6.2223\dots - \phi \cdot 6.2223\dots$ $x(1-6.2223\dots) = -\phi(6.2223\dots+1)$ $x = \boxed{\frac{-\phi(6.2223\dots+1)}{1-6.2223\dots}} = \boxed{2.2377\dots}$

If we plug this number into our equation, it does return with $2$ , so this is the solution! To get the answer, we can plug this number into the required calculation and we find that the answer is $\huge \green{\boxed{7}}$