If alpha, beta, gamma are the roots of x^3-2x | vedclass

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Question

(A) Given the cubic equation $x^3-2x^2+3x-4=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = 2$ $\alpha\beta+\bet

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(A) Given the cubic equation $x^3-2x^2+3x-4=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = 2$ $\alpha\beta+\beta\gamma+\gamma\alpha = 3$ $\alpha\beta\gamma = 4$ We need to find the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$. 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 $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)$. This simplifies to $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma)$. Substituting the known values: $(3)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(4)(2)$. $9 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 16$. $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 9 - 16 = -7$.

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+3x-4=0$,then the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ is

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