$\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ prove that $(x + y + 1) = A (1 - x - y - 2 x y)$ where $A$ is parameter.

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Solution

$\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$
$\frac{d y}{d x} = - \frac{y^{2} + y + 1}{x^{2} + x + 1}$
$\frac{d y}{y^{2} + y + 1} = - \frac{d x}{x^{2} + x + 1}$
$\frac{d y}{y^{2} + 2 \times \frac{1}{2} \times y + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} = - \frac{d x}{x^{2} + 2 \times \frac{1}{2} \times x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1}$
$\frac{d y}{{(y + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = - \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$
$\int \frac{d y}{{(y + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d y = - \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$
$\frac{2}{\sqrt{3}} {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = - \frac{2}{\sqrt{3}} {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + c$
$\frac{2}{\sqrt{3}} {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + \frac{2}{\sqrt{3}} {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = c$
$\frac{2}{\sqrt{3}} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = c$
$\frac{2}{\sqrt{3}} {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{2 y + 1}{\sqrt{3}} + \frac{2 x + 1}{\sqrt{3}}}{1 - \frac{2 y + 1}{\sqrt{3}} \times \frac{2 x + 1}{\sqrt{3}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = c$
$\frac{\frac{2 y + 1}{\sqrt{3}} + \frac{2 x + 1}{\sqrt{3}}}{1 - \frac{2 y + 1}{\sqrt{3}} \times \frac{2 x + 1}{\sqrt{3}}} = tan (\frac{\sqrt{3} c}{2})$
$\frac{\frac{2 y + 1}{\sqrt{3}} + \frac{2 x + 1}{\sqrt{3}}}{1 - \frac{2 y + 1}{\sqrt{3}} \times \frac{2 x + 1}{\sqrt{3}}} = c_{1}$ where $c_{1} = tan (\frac{\sqrt{3} c}{2})$
$\frac{2 x + 2 y + 2}{\sqrt{3}} = c_{1} (1 - \frac{(2 y + 1) (2 x + 1)}{3})$
$2 \sqrt{3} (x + y + 1) = c_{1} (3 - 4 x y - 2 x - 2 y - 1)$
$2 \sqrt{3} (x + y + 1) = c_{1} (2 - 4 x y - 2 x - 2 y)$
$2 \sqrt{3} (x + y + 1) = 2 c_{1} (1 - 2 x y - x - y)$
$\sqrt{3} (x + y + 1) = c_{1} (1 - 2 x y - x - y)$ is the required general solution.

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