Without expanding at any stage, prove that `|(a-b,1,a),(b-c,1,b),(c-a,1,c)| = |(a,1,b),(b,1,c),(c,1,a)|`

Answer

To prove that

\[ |(a-b, 1, a), (b-c, 1, b), (c-a, 1, c)| = |(a, 1, b), (b, 1, c), (c, 1, a)| \]

we will use properties of determinants without expanding them.

### Step 1: Rewrite the Determinant
Let's denote the left-hand side determinant as \( D_1 \) and the right-hand side determinant as \( D_2 \).

\[ D_1 = |(a-b, 1, a), (b-c, 1, b), (c-a, 1, c)| \]
\[ D_2 = |(a, 1, b), (b, 1, c), (c, 1, a)| \]

### Step 2: Apply Column Operations
We will perform column operations on \( D_1 \). Specifically, we will subtract the second column from the first column.

\[ D_1 = |(a-b-1, 1, a), (b-c-1, 1, b), (c-a-1, 1, c)| \]

### Step 3: Simplify the First Column
This simplifies to:

\[ D_1 = |(a-1-b, 1, a), (b-1-c, 1, b), (c-1-a, 1, c)| \]

### Step 4: Factor Out the Common Terms
Now, we can factor out \(-1\) from the first column:

\[ D_1 = -1 |(b-a, 1, a), (c-b, 1, b), (a-c, 1, c)| \]

### Step 5: Change the Order of Columns
Now, we can swap the first and second columns of the determinant, which introduces a negative sign:

\[ D_1 = -(-1) |(1, b-a, a), (1, c-b, b), (1, a-c, c)| \]

This gives us:

\[ D_1 = |(1, b-a, a), (1, c-b, b), (1, a-c, c)| \]

### Step 6: Rearranging the Columns
Now we can rearrange the columns to match the form of \( D_2 \):

\[ D_1 = |(a, 1, b), (b, 1, c), (c, 1, a)| \]

### Conclusion
Thus, we have shown that:

\[ |(a-b, 1, a), (b-c, 1, b), (c-a, 1, c)| = |(a, 1, b), (b, 1, c), (c, 1, a)| \]