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Question

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

Vedclass · Accepted Answer

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

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(A) Let the roots of the equation $x^3 + 3x^2 - 1 = 0$ be $\alpha, \beta,$ and $\gamma$.From Vieta's formulas,the product of the roots is $\alpha \beta \gamma = -(\frac{-1}{1}) = 1$.Therefore,$\alpha \beta = \frac{1}{\gamma}$.Let $y = \alpha \beta = \frac{1}{\gamma}$,which implies $\gamma = \frac{1}{y}$.Since $\gamma$ is a root of the original equation $x^3 + 3x^2 - 1 = 0$,we substitute $x = \gamma = \frac{1}{y}$ into the equation:$(\frac{1}{y})^3 + 3(\frac{1}{y})^2 - 1 = 0$.Multiplying the entire equation by $y^3$,we get:$1 + 3y - y^3 = 0$.Rearranging the terms,we get $y^3 - 3y - 1 = 0$.Replacing $y$ with $x$,the required equation is $x^3 - 3x - 1 = 0$.

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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