Show that `|[""^xC_r, ""^x C_(r+1),""^x C_(r+2)],[""^y C_r,""^y C_(r+1),""^y C_(r+2)],[""^z C_r,""^z C_(r+1),""^z C_(r+2)]|=|[""^(x)C_r, ""^(x+1) C_(r+1),""^(x+2) C_(r+2)],[""^y C_r,""^(y+1) C_(r+1),""^(y+2) C_(r+2)],[""^z C_r,""^(z+1) C_(r+1),""^(z+2) C_(r+2)]|`

Options

A. `0`
B. `2^(n)`
C. `"^(x+y+z)C_(r )`
D. `"^(x+y+z)C_(r +2)`

Explanation

To show that

\[ \begin{vmatrix} {^xC_r} & {^xC_{r+1}} & {^xC_{r+2}} \\ {^yC_r} & {^yC_{r+1}} & {^yC_{r+2}} \\ {^zC_r} & {^zC_{r+1}} & {^zC_{r+2}} \end{vmatrix} = \begin{vmatrix} {^{x}C_r} & {^{x+1}C_{r+1}} & {^{x+2}C_{r+2}} \\ {^{y}C_r} & {^{y+1}C_{r+1}} & {^{y+2}C_{r+2}} \\ {^{z}C_r} & {^{z+1}C_{r+1}} & {^{z+2}C_{r+2}} \end{vmatrix} \]

we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Start with the LHS
We begin with the determinant:

\[ D_1 = \begin{vmatrix} {^xC_r} & {^xC_{r+1}} & {^xC_{r+2}} \\ {^yC_r} & {^yC_{r+1}} & {^yC_{r+2}} \\ {^zC_r} & {^zC_{r+1}} & {^zC_{r+2}} \end{vmatrix} \]

### Step 2: Perform Column Operations
We will add the second column to the third column:

\[ D_1 = \begin{vmatrix} {^xC_r} & {^xC_{r+1}} & {^xC_{r+2} + ^xC_{r+1}} \\ {^yC_r} & {^yC_{r+1}} & {^yC_{r+2} + ^yC_{r+1}} \\ {^zC_r} & {^zC_{r+1}} & {^zC_{r+2} + ^zC_{r+1}} \end{vmatrix} \]

### Step 3: Apply the Binomial Coefficient Identity
Using the identity \( {^nC_r} + {^nC_{r-1}} = {^{n+1}C_r} \), we can simplify the third column:

\[ D_1 = \begin{vmatrix} {^xC_r} & {^xC_{r+1}} & {^{x+1}C_{r+2}} \\ {^yC_r} & {^yC_{r+1}} & {^{y+1}C_{r+2}} \\ {^zC_r} & {^zC_{r+1}} & {^{z+1}C_{r+2}} \end{vmatrix} \]

### Step 4: Perform Another Column Operation
Now we will add the first column to the second column:

\[ D_1 = \begin{vmatrix} {^xC_r} & {^{x+1}C_{r+1}} & {^{x+1}C_{r+2}} \\ {^yC_r} & {^{y+1}C_{r+1}} & {^{y+1}C_{r+2}} \\ {^zC_r} & {^{z+1}C_{r+1}} & {^{z+1}C_{r+2}} \end{vmatrix} \]

### Step 5: Apply the Binomial Coefficient Identity Again
Using the same identity again, we can simplify the second column:

\[ D_1 = \begin{vmatrix} {^xC_r} & {^{x+1}C_{r+1}} & {^{x+2}C_{r+2}} \\ {^yC_r} & {^{y+1}C_{r+1}} & {^{y+2}C_{r+2}} \\ {^zC_r} & {^{z+1}C_{r+1}} & {^{z+2}C_{r+2}} \end{vmatrix} \]

### Step 6: Conclusion
Now we have transformed the LHS into the form of the RHS:

\[ D_1 = \begin{vmatrix} {^{x}C_r} & {^{x+1}C_{r+1}} & {^{x+2}C_{r+2}} \\ {^{y}C_r} & {^{y+1}C_{r+1}} & {^{y+2}C_{r+2}} \\ {^{z}C_r} & {^{z+1}C_{r+1}} & {^{z+2}C_{r+2}} \end{vmatrix} \]

Thus, we have shown that:

\[ \begin{vmatrix} {^xC_r} & {^xC_{r+1}} & {^xC_{r+2}} \\ {^yC_r} & {^yC_{r+1}} & {^yC_{r+2}} \\ {^zC_r} & {^zC_{r+1}} & {^zC_{r+2}} \end{vmatrix} = \begin{vmatrix} {^{x}C_r} & {^{x+1}C_{r+1}} & {^{x+2}C_{r+2}} \\ {^{y}C_r} & {^{y+1}C_{r+1}} & {^{y+2}C_{r+2}} \\ {^{z}C_r} & {^{z+1}C_{r+1}} & {^{z+2}C_{r+2}} \end{vmatrix} \]