Find the non-zero roots of the equation.
`(i) Delta=|{:(,a,b,ax+b),(,b,c,bx+c),(,ax+b,bx+c,c):}|=0`

Answer

To find the non-zero roots of the equation given by the determinant of the matrix:

\[ \Delta = \begin{vmatrix} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & c \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant of the matrix:

\[ \Delta = \begin{vmatrix} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & c \end{vmatrix} \]

### Step 2: Apply column operations
We will perform column operations to simplify the determinant. Multiply the first column by \(x\):

\[ C_1 \rightarrow xC_1 \]

This modifies the matrix to:

\[ \begin{vmatrix} ax & b & ax + b \\ bx & c & bx + c \\ ax + b & bx + c & c \end{vmatrix} \]

### Step 3: Further simplify using column operations
Next, we will perform the operation \(C_3 \rightarrow C_3 - C_1 + C_2\):

\[ C_3 \rightarrow C_3 - C_1 + C_2 \]

This results in:

\[ \begin{vmatrix} ax & b & b \\ bx & c & c - b \\ ax + b & bx + c & 0 \end{vmatrix} \]

### Step 4: Expand the determinant
Now we can expand the determinant along the third column:

\[ \Delta = b \begin{vmatrix} ax & b \\ bx & c \end{vmatrix} + (c - b) \begin{vmatrix} ax & b \\ ax + b & bx + c \end{vmatrix} \]

### Step 5: Calculate the 2x2 determinants
Calculating the first determinant:

\[ \begin{vmatrix} ax & b \\ bx & c \end{vmatrix} = (ax)(c) - (b)(bx) = acx - b^2x \]

Calculating the second determinant:

\[ \begin{vmatrix} ax & b \\ ax + b & bx + c \end{vmatrix} = (ax)(bx + c) - (b)(ax + b) = abx^2 + acx - abx - b^2 = abx^2 + (ac - ab)x - b^2 \]

### Step 6: Substitute back into the determinant
Substituting back into the determinant gives:

\[ \Delta = b(acx - b^2x) + (c - b)(abx^2 + (ac - ab)x - b^2) \]

### Step 7: Set the determinant to zero
Setting \(\Delta = 0\):

\[ b(acx - b^2x) + (c - b)(abx^2 + (ac - ab)x - b^2) = 0 \]

### Step 8: Factor out common terms
This simplifies to:

\[ x \left( b(ac - b^2) + (c - b)(abx + (ac - ab)) \right) = 0 \]

### Step 9: Solve for non-zero roots
The non-zero roots can be found from:

\[ b(ac - b^2) + (c - b)(abx + (ac - ab)) = 0 \]

This leads to:

\[ abx + (ac - ab) = 0 \implies x = -\frac{(ac - ab)}{ab} \]

### Final Answer
The non-zero root is:

\[ x = -\frac{(ac - ab)}{ab} \]