The equation $$x^3 - \frac{1}{x} = 4$$ has two real roots $$\alpha, \beta$$. The value of $$(\alpha + \beta)^2$$ is

Multiplying by $$x$$ gives the quartic $$x^4 - 4x - 1 = 0$$, which factorises as $$\left(x^2 + \sqrt{2}x + 1 + \sqrt{2}\right)\left(x^2 - \sqrt{2}x + 1 - \sqrt{2}\right) = 0$$. The first factor has negative discriminant while the second has discriminant $$4\sqrt{2} - 2 > 0$$, so the two real roots come from $$x^2 - \sqrt{2}x + 1 - \sqrt{2} = 0$$. Thus $$\alpha + \beta = \sqrt{2}$$ and $$(\alpha+\beta)^2 = 2$$.