If `a+b+c!=` and `|(a,b,c),(b,c,a),(c,a,b)|=0` then prove that `a=b=c`

Explanation

To prove that \( a = b = c \) given that \( a + b + c \neq 0 \) and the determinant \( |(a,b,c),(b,c,a),(c,a,b)| = 0 \), we can follow these steps:

### Step 1: Write the determinant
The determinant can be expressed as:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Given that \( D = 0 \).

### Step 2: Apply row transformation
We can perform a row operation by transforming \( R_1 \) to \( R_1 + R_2 + R_3 \):
\[ D = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 3: Factor out \( a+b+c \)
We can factor out \( a+b+c \) from the first row:
\[ D = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]
Since \( D = 0 \), we have:
\[ (a+b+c) \cdot \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

### Step 4: Analyze the determinant
Since \( a + b + c \neq 0 \), we must have:
\[ \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

### Step 5: Perform column operations
Now, we will perform column operations to simplify the determinant:
1. Transform \( C_1 \) to \( C_1 - C_2 \) and \( C_2 \) to \( C_2 - C_3 \):
\[ \begin{vmatrix} 1 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} \]

### Step 6: Expand the determinant
Now, we can expand the determinant:
\[ D = 1 \cdot \begin{vmatrix} b-c & c-a \\ c-a & a-b \end{vmatrix} \]
Calculating this gives:
\[ (b-c)(a-b) - (c-a)(c-a) = 0 \]
This simplifies to:
\[ (b-c)(a-b) - (c-a)^2 = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ (b-c)(a-b) = (c-a)^2 \]

### Step 8: Analyze the implications
Since \( (c-a)^2 \geq 0 \), we can conclude that \( (b-c)(a-b) \) must also be non-negative.

### Step 9: Conclude the equality
The only way for both sides to be equal and non-negative is if:
1. \( b = c \)
2. \( a = b \)
3. \( a = c \)

Thus, we conclude:
\[ a = b = c \]

### Final Conclusion
Hence, we have proved that \( a = b = c \).

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