If Delta=left|begin{array}{ccc}x-2 & 2 x-3 & | vedclass

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(C) Given $\Delta = \left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \ 2 x-3 & 3 x-4 & 4 x-5 \ 3 x-5 & 5 x-8 & 10 x-17\end{array}ight| = Ax^{3}+Bx^{2}+

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$-3$

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(C) Given $\Delta = \left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \ 2 x-3 & 3 x-4 & 4 x-5 \ 3 x-5 & 5 x-8 & 10 x-17\end{array}ight| = Ax^{3}+Bx^{2}+Cx+D$. Applying row operations $R_{2} ightarrow R_{2}-R_{1}$ and $R_{3} ightarrow R_{3}-R_{2}$: $\Delta = \left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \ (2x-3)-(x-2) & (3x-4)-(2x-3) & (4x-5)-(3x-4) \ (3x-5)-(2x-3) & (5x-8)-(3x-4) & (10x-17)-(4x-5)\end{array}ight|$ $\Delta = \left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \ x-1 & x-1 & x-1 \ x-2 & 2x-4 & 6x-12\end{array}ight|$ Taking $(x-1)$ common from $R_{2}$ and $(x-2)$ common from $R_{3}$: $\Delta = (x-1)(x-2) \left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \ 1 & 1 & 1 \ 1 & 2 & 6\end{array}ight|$ Expanding the determinant: $\Delta = (x-1)(x-2) [ (x-2)(6-2) - (2x-3)(6-1) + (3x-4)(2-1) ]$ $\Delta = (x-1)(x-2) [ 4(x-2) - 5(2x-3) + 1(3x-4) ]$ $\Delta = (x-1)(x-2) [ 4x-8 - 10x+15 + 3x-4 ]$ $\Delta = (x-1)(x-2) [ -3x+3 ] = -3(x-1)(x-2)(x-1) = -3(x-1)^{2}(x-2)$ Expanding $-3(x^{2}-2x+1)(x-2) = -3(x^{3}-2x^{2}-2x^{2}+4x+x-2) = -3(x^{3}-4x^{2}+5x-2) = -3x^{3}+12x^{2}-15x+6$. Comparing with $Ax^{3}+Bx^{2}+Cx+D$,we get $B=12$ and $C=-15$. Therefore,$B+C = 12-15 = -3$.

If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=Ax^{3}+Bx^{2}+Cx+D$,then $B+C$ is equal to

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