If x = a + 2b satisfies the cubic equation (a | vedclass

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(B) Given the determinant $f(x) = \begin{vmatrix} a - x & b & b \ b & a - x & b \ b & b & a - x \end{vmatrix} = 0$. Applying the row operation $R_1

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real and coincident

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(B) Given the determinant $f(x) = \begin{vmatrix} a - x & b & b \ b & a - x & b \ b & b & a - x \end{vmatrix} = 0$. Applying the row operation $R_1 	o R_1 + R_2 + R_3$,we get: $f(x) = \begin{vmatrix} a - x + 2b & a - x + 2b & a - x + 2b \ b & a - x & b \ b & b & a - x \end{vmatrix} = 0$. Taking $(a - x + 2b)$ as a common factor from the first row: $f(x) = (a - x + 2b) \begin{vmatrix} 1 & 1 & 1 \ b & a - x & b \ b & b & a - x \end{vmatrix} = 0$. Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$: $f(x) = (a - x + 2b) \begin{vmatrix} 1 & 0 & 0 \ b & a - x - b & 0 \ b & 0 & a - x - b \end{vmatrix} = 0$. Expanding along the first row: $f(x) = (a - x + 2b)(a - x - b)^2 = 0$. The roots are $x = a + 2b$ and $x = a - b$ (twice). Since $a, b \in R$,the other two roots are $x = a - b$,which are real and coincident.

If $x = a + 2b$ satisfies the cubic equation $(a, b \in R)$ $f(x) = \begin{vmatrix} a - x & b & b \\ b & a - x & b \\ b & b & a - x \end{vmatrix} = 0$,then its other two roots are

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