Suppose alpha, beta, gamma are roots of x^3+x | vedclass

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(C) Given,the roots of $x^3+x^2+2x+3=0$ are $\alpha, \beta$ and $\gamma$.From Vieta's formulas:$\alpha+\beta+\gamma = -1$$\alpha\beta+\beta\gamma+\gam

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$x^3+2x^2+3x-1$

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(C) Given,the roots of $x^3+x^2+2x+3=0$ are $\alpha, \beta$ and $\gamma$.From Vieta's formulas:$\alpha+\beta+\gamma = -1$$\alpha\beta+\beta\gamma+\gamma\alpha = 2$$\alpha\beta\gamma = -3$Let the roots of $f(x)=0$ be $S_1 = \alpha+\beta$,$S_2 = \beta+\gamma$,$S_3 = \gamma+\alpha$.Note that $S_1 = -1-\gamma$,$S_2 = -1-\alpha$,$S_3 = -1-\beta$.The sum of roots of $f(x)$ is $S_1+S_2+S_3 = 2(\alpha+\beta+\gamma) = 2(-1) = -2$.The sum of roots taken two at a time is $S_1S_2+S_2S_3+S_3S_1 = (-1-\gamma)(-1-\alpha) + (-1-\alpha)(-1-\beta) + (-1-\beta)(-1-\gamma)$$= (1+\alpha+\gamma+\alpha\gamma) + (1+\alpha+\beta+\alpha\beta) + (1+\beta+\gamma+\beta\gamma)$$= 3 + 2(\alpha+\beta+\gamma) + (\alpha\beta+\beta\gamma+\gamma\alpha) = 3 + 2(-1) + 2 = 3$.The product of roots is $S_1S_2S_3 = (-1-\gamma)(-1-\alpha)(-1-\beta) = -((1+\gamma)(1+\alpha)(1+\beta))$$= -(1 + (\alpha+\beta+\gamma) + (\alpha\beta+\beta\gamma+\gamma\alpha) + \alpha\beta\gamma) = -(1 - 1 + 2 - 3) = -(-1) = 1$.Thus,$f(x) = x^3 - (	ext{sum of roots})x^2 + (	ext{sum of roots taken two at a time})x - (	ext{product of roots})$$f(x) = x^3 - (-2)x^2 + 3x - 1 = x^3+2x^2+3x-1$.

Suppose $\alpha, \beta, \gamma$ are roots of $x^3+x^2+2x+3=0$. If $f(x)=0$ is a cubic polynomial equation whose roots are $\alpha+\beta, \beta+\gamma, \gamma+\alpha$,then $f(x)$ is equal to

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