If the line y = sqrt{3}x cuts the curve x^3 + | vedclass

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(D) The line is $y = \sqrt{3}x$. This can be written in parametric form as $x = r \cos(	heta)$ and $y = r \sin(	heta)$,where $	an(	heta) = \sqrt{3

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(D) The line is $y = \sqrt{3}x$. This can be written in parametric form as $x = r \cos(	heta)$ and $y = r \sin(	heta)$,where $	an(	heta) = \sqrt{3}$,so $	heta = 60^\circ$. Thus,$x = r \cos(60^\circ) = \frac{r}{2}$ and $y = r \sin(60^\circ) = \frac{r\sqrt{3}}{2}$.Substituting these into the curve equation $x^3 + 3y^2 + 4x + 5y - 1 = 0$:$(\frac{r}{2})^3 + 3(\frac{r\sqrt{3}}{2})^2 + 4(\frac{r}{2}) + 5(\frac{r\sqrt{3}}{2}) - 1 = 0$$\frac{r^3}{8} + \frac{9r^2}{4} + 2r + \frac{5\sqrt{3}r}{2} - 1 = 0$Multiplying by $8$:$r^3 + 18r^2 + (16 + 20\sqrt{3})r - 8 = 0$Let the roots be $r_1, r_2, r_3$,which represent the distances $OA, OB, OC$. The product of the roots is given by the constant term with a sign change (for a cubic $ar^3 + br^2 + cr + d = 0$,product is $-d/a$):$OA \cdot OB \cdot OC = r_1 r_2 r_3 = -(-8)/1 = 8$.

If the line $y = \sqrt{3}x$ cuts the curve $x^3 + 3y^2 + 4x + 5y - 1 = 0$ at the points $A, B, C$,then the product $OA \cdot OB \cdot OC$ is

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