If alpha, beta, gamma are the roots of the eq | vedclass

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(A) Given the equation $x^3 + x^2 + x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.By Vieta's formulas:$\alpha + \beta + \gamma = -1$$\alpha \beta +

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$(i)$ $\rightarrow$ $A$,(ii) $\rightarrow$ $A$,(iii) $\rightarrow$ $D$,(iv) $\rightarrow$ $B$

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(A) Given the equation $x^3 + x^2 + x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.By Vieta's formulas:$\alpha + \beta + \gamma = -1$$\alpha \beta + \beta \gamma + \gamma \alpha = 1$$\alpha \beta \gamma = -1$$(i)$ $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha \beta + \beta \gamma + \gamma \alpha}{\alpha \beta \gamma} = \frac{1}{-1} = -1$. Thus,$(i)$ $\rightarrow$ $A$.(ii) $\alpha^3 + \beta^3 + \gamma^3$: Since $\alpha, \beta, \gamma$ are roots,$\alpha^3 = -\alpha^2 - \alpha - 1$,$\beta^3 = -\beta^2 - \beta - 1$,$\gamma^3 = -\gamma^2 - \gamma - 1$.Summing these: $\alpha^3 + \beta^3 + \gamma^3 = -(\alpha^2 + \beta^2 + \gamma^2) - (\alpha + \beta + \gamma) - 3$.We know $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = (-1)^2 - 2(1) = 1 - 2 = -1$.So,$\alpha^3 + \beta^3 + \gamma^3 = -(-1) - (-1) - 3 = 1 + 1 - 3 = -1$. Thus,(ii) $\rightarrow$ $A$.(iii) $\alpha^4 + \beta^4 + \gamma^4$: Multiply the original equation by $x$: $x^4 + x^3 + x^2 + x = 0$.Summing for roots: $(\alpha^4 + \beta^4 + \gamma^4) + (\alpha^3 + \beta^3 + \gamma^3) + (\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) = 0$.$(\alpha^4 + \beta^4 + \gamma^4) + (-1) + (-1) + (-1) = 0 \Rightarrow \alpha^4 + \beta^4 + \gamma^4 = 3$. Thus,(iii) $\rightarrow$ $D$.(iv) $(\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 2(\alpha^2 + \beta^2 + \gamma^2) - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 2(-1) - 2(1) = -2 - 2 = -4$. Thus,(iv) $\rightarrow$ $B$.The correct match is $(i)$ $\rightarrow$ $A$,(ii) $\rightarrow$ $A$,(iii) $\rightarrow$ $D$,(iv) $\rightarrow$ $B$.

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