Product of Complex Solutions

$$ Let $y=x+\dfrac{1}{x}$ , then $$ $\begin{aligned} \left(x+\frac{1}{x}\right)^3&=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\\ y^3&=52+3y\\ y^3-3y-52&=0\\ (y-4)(y^2+4y+13)&=0. \end{aligned}$ $$ We obtain $\,y_1=4$ and $\begin{aligned} y^2+4y+13&=0\\ (y+2)^2+9&=0\\ (y+2)^2&=-9\\ y+2&=\sqrt{-9}\\ y+2&=\pm3i\\ y&=-2\pm3i\\ y_2=-2+3i\;&\text{or}\;\;y_3=-2-3i \end{aligned}$ $$ Since we only need the complex roots, therefore the product of all distinct complex values of $x+\dfrac{1}{x}$ is $$ $\begin{aligned} y_2y_3&=(-2+3i)(-2-3i)\\ \left(x_2+\frac{1}{x_2}\right)\left(x_3+\frac{1}{x_3}\right)&=(-2)^2-(3i)^2\\ &=4+9\\ &=\boxed{13} \end{aligned}$ $$ It has been proven WITHOUT using Wolfram Alpha!! :P $$

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Let $a= x+\dfrac{1}{x}$

$(x+\dfrac{1}{x})^{3} = x^{3}+\dfrac{1}{x^{3}}+3(x+\dfrac{1}{x})$

$x^{3}+\dfrac{1}{x^{3}}=(x+\dfrac{1}{x})^{3}-3(x+\dfrac{1}{x})=a^{3}-3a=52$

$a^{3}-3a-52=(a-4)(a^{2}+4a+13)$

Since 4 does not satisfy the required condition , we will find the roots of the quadratic.

Discriminant of the quadratic equation= $4^{2}-4(13)(1)=-36 < 0$

Therefore,the given equation has both complex roots. Using Vieta's formula , product = $\boxed{13}$