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

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(B) Given the cubic equation $x^3-3x^2+3x+1=0$.By Vieta's formulas,we have:$\alpha+\beta+\gamma = 3$$\alpha\beta+\beta\gamma+\gamma\alpha = 3$$\alpha\

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Given the cubic equation $x^3-3x^2+3x+1=0$.By Vieta's formulas,we have:$\alpha+\beta+\gamma = 3$$\alpha\beta+\beta\gamma+\gamma\alpha = 3$$\alpha\beta\gamma = -1$We need to find $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$.Using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$,let $a=\alpha\beta, b=\beta\gamma, c=\gamma\alpha$.Then $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)$.Factor out $\alpha\beta\gamma$ from the second term:$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\beta+\alpha+\gamma)$.Substitute the known values:$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (3)^2 - 2(-1)(3) = 9 + 6 = 15$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-3x^2+3x+1=0$,then $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=$

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