Square on the curve A square is drawn on the curve $y = x^3 + 27\cdot x^2 + 8\cdot x + 91$. What would be the area of the square. All the points of the square should lie on the curve.
All four points lie on the curve (not the sides)

Time for some reverse engineering.  Suppose we start with a square; translate it so that it is centred at the origin.  Then for some $s$ and $\theta$, the vertices are $(s \cos(\theta), s \sin(\theta))$, $(s \sin(\theta), -s \cos(\theta))$, $(-s \cos(\theta), -s \sin(\theta))$ and $(-s \sin(\theta), s \cos(\theta))$, and the area is $2 s^2$.  The cubic that interpolates these points is $$f(x) = \frac{4}{s^2 \sin(4 \theta)} x^3 - \frac{3 + \cos(4 \theta)}{\sin(4 \theta)} x$$
Note that for $b \ge \sqrt{8}$, $b = (3 +  \cos(4 \theta))/\sin(4 \theta)$ has  real solutions, namely $$\sin(4 \theta) = \frac{3b \pm \sqrt{b^2-8}}{b^2  - 1}$$ Note also that  these solutions have $\sin(4 \theta) > 0$.  We can then take $s=2/\sqrt{\sin(4\theta)}$ to make the cubic $f(x) = x^3 - b x$, and the area is 
$$2s^2 = \frac{8}{\sin(4\theta)} = 3 b \mp \sqrt{b^2 - 8}$$  The given problem has $b = 235$  and area $705 \mp \sqrt{55217}$.
