Let a, b, c be such that b(a+c) neq 0. If lef | vedclass

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(D) Let the given equation be $D_1 + D_2 = 0$. We have $D_1 = \left|\begin{array}{ccc}a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1\end{array}igh

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any odd integer

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(D) Let the given equation be $D_1 + D_2 = 0$. We have $D_1 = \left|\begin{array}{ccc}a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1\end{array}ight|$. Taking the transpose of $D_2$,we get $D_2 = \left|\begin{array}{ccc}a+1 & a-1 & (-1)^{n+2} a \ b+1 & b-1 & (-1)^{n+1} b \ c-1 & c+1 & (-1)^{n} c\end{array}ight|$. Now,swapping the first and third columns of $D_2$ twice (or rearranging columns) to match the structure of $D_1$: $D_2 = \left|\begin{array}{ccc}(-1)^{n+2} a & a+1 & a-1 \ (-1)^{n+1} b & b+1 & b-1 \ (-1)^{n} c & c-1 & c+1\end{array}ight|$. Adding $D_1$ and $D_2$,we get: $\left|\begin{array}{ccc}a(1 + (-1)^{n+2}) & a+1 & a-1 \ b(-1 + (-1)^{n+1}) & b+1 & b-1 \ c(1 + (-1)^{n}) & c-1 & c+1\end{array}ight| = 0$. For the determinant to be zero for arbitrary $a, b, c$,the first column must be zero. $1 + (-1)^{n+2} = 0 \Rightarrow (-1)^{n+2} = -1$,which implies $n+2$ is odd,so $n$ is odd. $-1 + (-1)^{n+1} = 0 \Rightarrow (-1)^{n+1} = 1$,which implies $n+1$ is even,so $n$ is odd. $1 + (-1)^{n} = 0 \Rightarrow (-1)^{n} = -1$,which implies $n$ is odd. Thus,$n$ is any odd integer.

Let $a, b, c$ be such that $b(a+c) \neq 0$. If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right| + \left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$,then the value of $n$ is:

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