The more it is simple,the more it is maths.

The function $f(x)$ has only two roots when its local minimum value $f_{min}(x) = 0$ and only one root when $f_{min}(x) > 0$ .

Local maximum and minimum occur when $f'(x) = 0$ or $3x^2 - n = 0$ $\implies x = \pm \sqrt{\frac n3}$ . The local minimum is $f \left( \sqrt{\frac n3} \right)$ because $f''\left(\sqrt{\frac n3} \right) > 0$ . $\implies f \left( \sqrt{\frac a3} \right) = 0$ . Now, we have:

$\begin{aligned} f \left( \sqrt{\frac a3} \right) & = 0 \\ \implies \left( \sqrt{\frac a3} \right)^3 - a \left( \sqrt{\frac a3} \right) + 1 & = 0 \\ \frac a3 \left( \sqrt{\frac a3} \right) - a \left( \sqrt{\frac a3} \right) + 1 & = 0 \\ \frac 23 \cdot a \cdot \sqrt{\frac a3} & = 1 \\ a^\frac 32 & = \frac {3^\frac 32}2 \\ a & = \frac 3{\sqrt [3] 4} \approx \boxed{1.889} \end{aligned}$