<s>[INST] 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)
 [/INST] 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 $705 \mp \sqrt{55217}$.
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