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

We have to evaluate the determinant

$$ \Delta \;=\; \begin{vmatrix} x-2 & 2x-3 & 3x-4\\ 2x-3 & 3x-4 & 4x-5\\ 3x-5 & 5x-8 & 10x-17 \end{vmatrix} \;=\; Ax^{3}+Bx^{2}+Cx+D, $$

and then find the value of $$B+C$$.

First, to simplify the computation, we carry out elementary row operations. Remember that replacing a row by “that row minus another row” does not change the value of the determinant.

We perform the following operations:

$$ R_{2}\;\longrightarrow\; R_{2}-R_{1},\qquad R_{3}\;\longrightarrow\; R_{3}-R_{2}. $$

Let us calculate the new rows one by one.

For the second row:

$$$ \begin{aligned} (2x-3)-(x-2) &= x-1,\\ (3x-4)-(2x-3) &= x-1,\\ (4x-5)-(3x-4) &= x-1. \end{aligned} $$$ So the new second row is $$[\,x-1,\;x-1,\;x-1\,].$$

For the third row:

$$$ \begin{aligned} (3x-5)-(2x-3) &= x-2,\\ (5x-8)-(3x-4) &= 2x-4,\\ (10x-17)-(4x-5) &= 6x-12. \end{aligned} $$$ Hence the new third row is $$[\,x-2,\;2x-4,\;6x-12\,].$$

After these operations the determinant becomes

$$ \Delta =\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ x-1 & x-1 & x-1\\ x-2 & 2x-4 & 6x-12 \end{vmatrix}. $$

Now we observe common factors in rows:

• The entire second row contains a factor of $$x-1$$, because every entry is exactly $$x-1$$.
• The entire third row contains a factor of $$x-2$$, since $$x-2,\;2x-4=2(x-2),\;6x-12=6(x-2).$$

Factoring these out of the determinant (factor from a row comes out as a multiplicative factor) we get

$$ \Delta \;=\; (x-1)(x-2) \begin{vmatrix} \,x-2 & 2x-3 & 3x-4\\ 1 & 1 & 1\\ 1 & 2 & 6 \end{vmatrix}. $$

Let us denote the remaining 3 × 3 determinant by $$M$$. We now evaluate

$$ M \;=\; \begin{vmatrix} a & b & c\\ 1 & 1 & 1\\ 1 & 2 & 6 \end{vmatrix}, $$ where for convenience we have written $$a=x-2,\quad b=2x-3,\quad c=3x-4.$$

We shall expand this determinant along the first row, using the usual cofactor formula $$ \begin{vmatrix} p_{11}&p_{12}&p_{13}\\ p_{21}&p_{22}&p_{23}\\ p_{31}&p_{32}&p_{33} \end{vmatrix} = p_{11}(p_{22}p_{33}-p_{23}p_{32}) -p_{12}(p_{21}p_{33}-p_{23}p_{31}) +p_{13}(p_{21}p_{32}-p_{22}p_{31}). $$

Applying this rule, we obtain

$$$ \begin{aligned} M &= a\,(1\cdot6-1\cdot2)\;-\;b\,(1\cdot6-1\cdot1)\;+\;c\,(1\cdot2-1\cdot1)\\ &= a\,(6-2)\;-\;b\,(6-1)\;+\;c\,(2-1)\\ &= 4a\;-\;5b\;+\;c. \end{aligned} $$$

Now we substitute back the expressions for $$a,b,c$$:

$$$ \begin{aligned} 4a &= 4(x-2) = 4x-8,\\ -5b &= -5(2x-3) = -10x+15,\\ c &= 3x-4. \end{aligned} $$$ Adding these three quantities gives $$$ M = (4x-8) + (-10x+15) + (3x-4) = (4x-10x+3x) + (-8+15-4) = -3x + 3. $$$ Thus $$$ M = -3x+3 = -3(x-1). $$$

Putting this back into $$\Delta$$ we have

$$ \Delta = (x-1)(x-2)\,[-3(x-1)] = -3\,(x-1)^{2}(x-2). $$

Next we expand the cubic polynomial $$ (x-1)^{2}(x-2) $$ step by step. First,

$$$ (x-1)^{2} = x^{2}-2x+1. $$$ Multiplying by $$(x-2)$$ we get $$$ (x^{2}-2x+1)(x-2) = x^{3}-2x^{2} + x - 2x^{2}+4x-2 = x^{3}-4x^{2}+5x-2. $$$

Therefore

$$ \Delta = -3\bigl(x^{3}-4x^{2}+5x-2\bigr) = -3x^{3}+12x^{2}-15x+6. $$

Comparing with the standard form $$\Delta=Ax^{3}+Bx^{2}+Cx+D$$ we read off

$$ A=-3,\quad B=12,\quad C=-15,\quad D=6. $$

Finally, we compute

$$ B+C = 12 + (-15) = -3. $$

Hence, the correct answer is Option C.