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

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(C) Given the cubic equation $x^3-3x^2-4x+12=0$. Factoring the equation: $x^2(x-3)-4(x-3)=0$ $(x^2-4)(x-3)=0$ $(x-2)(x+2)(x-3)=0$ Thus,the roots are $

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$26$

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(C) Given the cubic equation $x^3-3x^2-4x+12=0$. Factoring the equation: $x^2(x-3)-4(x-3)=0$ $(x^2-4)(x-3)=0$ $(x-2)(x+2)(x-3)=0$ Thus,the roots are $\alpha=-2, \beta=2, \gamma=3$. We need to calculate $\sum(\alpha+\beta)^2 = (\alpha+\beta)^2 + (\beta+\gamma)^2 + (\gamma+\alpha)^2$. Substituting the values: $(\alpha+\beta)^2 = (-2+2)^2 = 0^2 = 0$ $(\beta+\gamma)^2 = (2+3)^2 = 5^2 = 25$ $(\gamma+\alpha)^2 = (3-2)^2 = 1^2 = 1$ Summing these values: $0 + 25 + 1 = 26$.

If $\alpha, \beta, \gamma$ are the roots of $x^3-3x^2-4x+12=0$,then $\sum(\alpha+\beta)^2$ is equal to

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