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(A) Let the three roots of the equation $x^3 + 3x^2 - 1 = 0$ be $\alpha, \beta,$ and $\gamma$.From the properties of roots,the product of the roots is

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$x^3 - 3x - 1 = 0$

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(A) Let the three roots of the equation $x^3 + 3x^2 - 1 = 0$ be $\alpha, \beta,$ and $\gamma$.From the properties of roots,the product of the roots is $\alpha \beta \gamma = -(\frac{-1}{1}) = 1$.Thus,$\alpha \beta = \frac{1}{\gamma}$.Since $\gamma$ is a root of the original equation,it satisfies $\gamma^3 + 3\gamma^2 - 1 = 0$.Let $x = \alpha \beta = \frac{1}{\gamma}$,which implies $\gamma = \frac{1}{x}$.Substituting $\gamma = \frac{1}{x}$ into the original equation:$(\frac{1}{x})^3 + 3(\frac{1}{x})^2 - 1 = 0$$\frac{1}{x^3} + \frac{3}{x^2} - 1 = 0$Multiplying by $x^3$,we get $1 + 3x - x^3 = 0$,which simplifies to $x^3 - 3x - 1 = 0$.

Let $\alpha$ and $\beta$ be two distinct roots of the equation $x^3 + 3x^2 - 1 = 0$. The equation which has $(\alpha \beta)$ as its root is equal to

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