Let alpha be a root of the equation (a-c)x^2 | vedclass

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(B) Given the matrix is singular,its determinant is $0$:$\Delta = \begin{vmatrix} \alpha^2 & \alpha & 1 \ 1 & 1 & 1 \ a & b & c \end{vmatrix} = 0$Ex

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$3$

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(B) Given the matrix is singular,its determinant is $0$:$\Delta = \begin{vmatrix} \alpha^2 & \alpha & 1 \ 1 & 1 & 1 \ a & b & c \end{vmatrix} = 0$Expanding along the first row:$\alpha^2(c-b) - \alpha(c-a) + (b-a) = 0$This is the same as the given equation $(a-c)x^2 + (b-a)x + (c-b) = 0$ if we note that $\alpha=1$ is a root because $(a-c) + (b-a) + (c-b) = 0$.Let $X = a-c$,$Y = b-a$,and $Z = c-b$. Note that $X+Y+Z = 0$.The expression is $\frac{X^2}{YZ} + \frac{Y^2}{XZ} + \frac{Z^2}{XY} = \frac{X^3 + Y^3 + Z^3}{XYZ}$.Since $X+Y+Z = 0$,we have the identity $X^3 + Y^3 + Z^3 = 3XYZ$.Therefore,the expression becomes $\frac{3XYZ}{XYZ} = 3$.

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