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$1+B y^3+C y^2 = 0$ $y = 1/3 \left( \frac{(-27 B^2 + 3 \sqrt{3} \sqrt{(27 B^4+4 B^2 C^3)} - 2 C^3)^{1/3}}{(2^{1/3} B)} + \frac{(2^{(1/3)} C^2)}{(B (-27 B^2 + 3 \sqrt{3} \sqrt{(27 B^4+4 B^2 C^3)} - 2 C^3)^{1/3})} - C/B \right)$ $y = -\frac{((1-i \sqrt{3}) (-27 B^2+3 \sqrt{3} \sqrt{27 B^4 + 4 B^2 C^3} - 2 C^3)^{1/3})} {(6 (2^{1/3}) B)} - \frac{(1+i \sqrt{3}) C^2} {(3 (2^{2/3}) B (-27 B^2 + 3 \sqrt{3} \sqrt{27 B^4+4 B^2 C^3}-2 C^3)^{1/3})} - \frac{C}{3 B}$ $y = -\frac{((1+i \sqrt{3}) (-27 B^2+3 \sqrt{3} \sqrt{27 B^4+4 B^2 C^3} - 2 C^3)^{1/3}) } { (6 (2^{1/3}) B)} -\frac{((1-i \sqrt{3}) C^2)} {(3 (2^{2/3}) B (-27 B^2+3 \sqrt{3} \sqrt{27 B^4+4 B^2 C^3}-2 C^3)^{1/3})} - \frac{C}{3 B}$ $D = \left( -27 B^2 + 3 \sqrt{3} \sqrt{(27 B^4+4 B^2 C^3)} - 2 C^3 \right)^{1/3}$ $y = 1/3 \left( \frac{D}{(2^{1/3} B)} + \frac{(2^{(1/3)} C^2)}{(B D)} - C/B \right)$ $y = -\frac{(1-i \sqrt{3}) D}{6 (2^{1/3}) B} - \frac{(1+i \sqrt{3}) C^2}{(3 (2^{2/3}) B D)} - \frac{C}{3 B}$ $y = -\frac{(1+i \sqrt{3}) D}{6 (2^{1/3}) B} -\frac{((1-i \sqrt{3}) C^2)}{(3 (2^{2/3}) B D)} - \frac{C}{3 B}$ $\frac{\partial y (B, C)}{\partial C} = - \frac{y}{2 C + 3 B y}$ $\frac{\partial y (B, C)}{\partial B} = - \frac{y^2}{2 C + 3 B y}$

Charles Rymal

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