Cubics, Rectangles and Parallelograms.

$m(x) = x(x - p)(x + p) = x^3 - p^2x \implies \dfrac{dm}{dx}|_{x = x_{0}} = 3x_{0}^2 - p^2 = \dfrac{x_{0}(x_{0} - p)(x_{0} + p)}{x_{0} - p} \implies$
$3x_{0}^2 - p^2 = x_{0} + px_{0} \implies 2x_{0}^2 - px_{0} - p^2 = 0 \implies x_{0} = p, -\dfrac{p}{2}$

$x_{0} \neq p \implies x_{0} = -\dfrac{p}{2} \implies B:(-\dfrac{p}{2}, \dfrac{3p^3}{8})$ and $D:(\dfrac{p}{2}, -\dfrac{3p^3}{8})$ .

$AC = BD = 2p = \dfrac{p}{4}\sqrt{16 + 9p^4} \implies 64 = 16 + 9p^4 \implies p^4 = \dfrac{16}{3} \implies p = \dfrac{2}{\sqrt[4]{3}}.$

To obtain the area of parallelogram $BFED$ you can:

(1) Find the area of the region bounded by the line the curve $m(x) = x^3 - \dfrac{4}{\sqrt{3}}$ and the line $y = -\dfrac{1}{\sqrt{3}}x - \dfrac{2}{\sqrt[4]{27}}$ then double the result.

Or

(2) Use slopes and distances to find the area.

Using (1):

$A(-\dfrac{2}{\sqrt[4]{3}}, 0), D(\dfrac{1}{\sqrt[4]{3}}, -\sqrt[4]{3})$

$\implies m_{AD} = -\dfrac{1}{\sqrt{3}} \implies y = -\dfrac{1}{\sqrt{3}}x - \dfrac{2}{\sqrt[4]{27}} = n(x)$

and,

$m(x) = x^3 - \dfrac{4}{\sqrt{3}}x$

$\implies A_{1} = A_{2} = \displaystyle\int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} m(x) - n(x) dx =$

$\displaystyle\int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} (x^3 - \sqrt{3}x + \dfrac{2}{3^{\frac{3}{4}}}) dx =\dfrac{x^4}{4} - \dfrac{\sqrt{3}}{2}x^2 + \dfrac{2}{3^{\frac{3}{4}}}x|_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} = \dfrac{4}{3}$

$\implies A_{BFED} = \dfrac{8}{3} = \dfrac{a}{b} \implies a + b =\boxed{11}.$

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Using (2) :

$m_{AD} = m_{BC} = -\dfrac{1}{\sqrt{3}}$ .

For $\overline{AD}: y = -\dfrac{1}{\sqrt{3}}x - \dfrac{2}{\sqrt[4]{27}}$

For $\perp$ line $FG$ that passes thru $F: (\dfrac{1}{\sqrt[4]{3}}, \dfrac{1}{\sqrt[4]{27}})$ :

$m_{\perp} = \sqrt{3} \implies y = \sqrt{3}x - \dfrac{2}{\sqrt[4]{27}} \implies \sqrt{3}x - \dfrac{2}{\sqrt[4]{27}} = -\dfrac{1}{\sqrt{3}}x - \dfrac{2}{\sqrt[4]{27}} \implies x = 0 \implies y = -\dfrac{2}{\sqrt[4]{27}}$

$\implies \overline{GF} = \dfrac{2}{\sqrt[4]{27}}$ and $ED = BF = \dfrac{4}{\sqrt[4]{27}} \implies$

$A_{BFED} = \dfrac{2}{3^{\frac{1}{4}}} * \dfrac{4}{3^{\frac{3}{4}}} = \dfrac{8}{3} = \dfrac{a}{b} \implies a + b = \boxed{11}$ .