$\Delta =\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ 2x-3& 3x-4& 4x-5\\ 3x-5& 5x-8& 10x-17\end{array}\right|$$=A{x}^3+B{x}^2+Cx+D$. Then $B+C=\_\_\_\_\_\_$

$c_3\to c_3-c_2,c_2\to c_2-c_1$gfdf

$\Delta =\left|\begin{array}{ccc}x-2& x-1& x-1\\ 2x-3& x-1& x-1\\ 3x-5& 2x-3& 5x-9\end{array}\right|$

$R_2\to R_2-R_1$

$\Delta =\left|\begin{array}{ccc}x-2& x-1& x-1\\ x-1& 0& 0\\ 3x-5& 2x-3& 5x-9\end{array}\right|$

$\Delta =-(x-1)\left[(x-1)(5x-9)\right]+(x-1)\left[(x-1)(2x-3)\right]$

$-(x-1)[(5x^2-14x+9)-(2x^2-5x+3)]$

$=-3x^3+12x^2-15x+6$

So, B+C=12-15= -3