# Prove the following

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Question:

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

then $B+C$ is equal to:

1. (1) $-1$

2. (2) 1

3. (3) $-3$

4. (4) 9

Correct Option: , 3

Solution:

$\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|$

$\Rightarrow \Delta=\left|\begin{array}{ccc}x-2 & x-1 & x-1 \\ 2 x-3 & x-1 & x-1 \\ 3 x-5 & 2 x-3 & 5 x-9\end{array}\right| \quad\left[\begin{array}{l}C_{3} \rightarrow C_{3}-C_{2} \\ C_{2} \rightarrow C_{2}-C_{1}\end{array}\right]$

$\Rightarrow \Delta=\left|\begin{array}{ccc}x-2 & x-1 & x-1 \\ x-1 & 0 & 0 \\ 3 x-5 & 2 x-3 & 5 x-9\end{array}\right| \quad\left[R_{2} \rightarrow R_{2}-R_{1}\right]$

$\Rightarrow \Delta=-(x-1)[(x-1)(5 x-9)-(x-1)(2 x-3)]$

$\Rightarrow \Delta=-(x-1)\left[\left(5 x^{2}-14 x+9\right)-\left(2 x^{2}-5 x+3\right)\right]$

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

So, $B+C=-3$