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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)

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    I think that it isn't possible to draw such square....Can you provide drawing ?2012-02-19
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    that is the problem i m also facing..but i have got this problem in my geometry problem sheet and the answer is non-zero2012-02-19
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    Hm, you can fit a square in the graph in $\mathbb R^4$ of $y=x^3+27x^2+8x+91$, but presumably that's not what the problem-sheet author seeks.2012-02-19
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    not in R4 ...we are talking about R22012-02-19
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    It is the same thing as looking for a square whose vertices lie on the graph $Y=X^3-235X$. I would look for square whose center is at $X=0$, $Y=0$. There seems to be a solution, but the computations are quite cumbersome; what answer is provided?2012-02-19
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    I would encourage the PO to write a more accurate description since it is not possible that "all four points line on the curve".2012-02-20

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