Employ Bezout's method to eliminate $x, y$ from $a x^{3} + b x^{2} y + c x y^{2} + d y^{3} = 0, a^{'} x^{3} + b x^{2} y + c^{'} x y^{2} + d^{'} y^{3} = 0$ .

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Solution

Dividing the two equations by $y^{3}$ and substituting $\frac{x}{y} = z$ , we get

$a z^{3} + b z^{2} + c z + d = 0, and a^{'} z^{3} + b^{'} z^{2} + c^{'} z + d^{'} = 0$

For the above equations, we have

$\frac{a}{a^{'}} = \frac{b z^{2} + c z + d}{b^{'} z^{2} + c^{'} z + d^{'}}; \frac{a z + b}{a^{'} z + b^{'}} = \frac{c z + d}{c^{'} z + d^{'}}; \frac{a z^{2} + b z + c}{a^{'} z^{2} + b^{'} z + c} = \frac{d}{d^{'}}$

Multiplying the above 3 equations, we get

$(a b^{'} - a b^{'}) z^{2} + (a c^{'} - a^{'} c) z + (a d^{'} - a d^{'}) = 0 \dots \dots (1)$
$(a c^{'} - a^{'} c) z^{2} + (a d^{'} - a^{'} d + b c^{'} - b^{'} c) z + (b d^{'} - b^{'} d) = 0 \dots \dots (2)$
$(a d^{'} - a^{'} d) z^{2} + (b d^{'} - b^{'} d) z + (c d^{'} - c^{'} d) = 0 \dots \dots (3)$

Eliminating $z^{2}$ and $z$ from the equation (1), (2) and (3), we have the eliminant as

$∣ ∣ ∣ ∣ ∣ \begin{matrix} (a b^{'} - a b^{'}) & (a c^{'} - a^{'} c) & (a d^{'} - a d^{'}) (a c^{'} - a^{'} c) & (a d^{'} - a^{'} d + b c^{'} - b^{'} c) & (b d^{'} - b^{'} d) (a d^{'} - a^{'} d) & (b d^{'} - b^{'} d) & (c d^{'} - c^{'} d) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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