If alpha, beta, gamma are the roots of left|b | vedclass

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(C) To find the roots $\alpha, \beta, \gamma$,we expand the determinant: $\left|\begin{array}{ccc} 1-x & -2 & 1 \ -2 & 4-x & -2 \ 1 & -2 & 1-x \end{

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(C) To find the roots $\alpha, \beta, \gamma$,we expand the determinant: $\left|\begin{array}{ccc} 1-x & -2 & 1 \ -2 & 4-x & -2 \ 1 & -2 & 1-x \end{array}ight|=0$. Expanding along the first row: $(1-x)[(4-x)(1-x) - 4] - (-2)[-2(1-x) - (-2)] + 1[4 - (4-x)] = 0$. $(1-x)[4 - 4x - x + x^2 - 4] + 2[-2 + 2x + 2] + 1[4 - 4 + x] = 0$. $(1-x)[x^2 - 5x] + 2[2x] + x = 0$. $x^2 - 5x - x^3 + 5x^2 + 4x + x = 0$. $-x^3 + 6x^2 = 0$,which implies $x^3 - 6x^2 = 0$. Comparing this with the cubic equation $ax^3 + bx^2 + cx + d = 0$,we have $a=1, b=-6, c=0, d=0$. For a cubic equation,the sum of roots taken two at a time is given by $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$. Substituting the values,$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{0}{1} = 0$.

If $\alpha, \beta, \gamma$ are the roots of $\left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0$,then $\alpha \beta+\beta \gamma+\gamma \alpha=$

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