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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

Vedclass · Accepted Answer

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

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

$A. \alpha + \beta + \gamma$	$(1) -\frac{43}{4}$
$B. \alpha^2 + \beta^2 + \gamma^2$	$(2) -\frac{7}{8}$
$C. \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$	$(3) 86$
$D. \frac{\alpha}{\beta \gamma} + \frac{\beta}{\gamma \alpha} + \frac{\gamma}{\alpha \beta}$	$(4) 0$
	$(5) 10$

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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