Is `((1)/(a)+(1)/(b)+(1)/(c )-1)` a factor of `|{:(1+a,1,1),(1,1+b,1),(1,1,1+c):}|` ?

Answer

To determine whether \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\) is a factor of the determinant

\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

we will follow these steps:

### Step 1: Write the determinant
The given determinant is

\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

### Step 2: Apply row operations
We will perform row operations to simplify the determinant. We can subtract the first row from the second and third rows:

- \(R_2 \to R_2 - R_1\)
- \(R_3 \to R_3 - R_1\)

This gives us:

\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 - (1 + a) & (1 + b) - 1 & 1 - 1 \\ 1 - (1 + a) & 1 - 1 & (1 + c) - 1 \end{vmatrix} \]

Simplifying this results in:

\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix} \]

### Step 3: Expand the determinant
Now we will expand the determinant along the first row:

\[ D = (1 + a) \begin{vmatrix} b & 0 \\ 0 & c \end{vmatrix} - 1 \begin{vmatrix} -a & 0 \\ -a & c \end{vmatrix} + 1 \begin{vmatrix} -a & b \\ -a & 0 \end{vmatrix} \]

Calculating these 2x2 determinants:

1. \(\begin{vmatrix} b & 0 \\ 0 & c \end{vmatrix} = bc\)

2. \(\begin{vmatrix} -a & 0 \\ -a & c \end{vmatrix} = -ac\)

3. \(\begin{vmatrix} -a & b \\ -a & 0 \end{vmatrix} = ab\)

Substituting these back into the determinant expansion gives:

\[ D = (1 + a)bc - (-ac) + ab \]

### Step 4: Simplify the expression
Now simplifying:

\[ D = (1 + a)bc + ac + ab \]

Expanding \((1 + a)bc\):

\[ D = bc + abc + ac + ab \]

Rearranging gives:

\[ D = abc + ab + ac + bc \]

### Step 5: Factor the expression
Now, we can factor out common terms:

\[ D = c(ab + b) + ab \]

### Step 6: Check if \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\) is a factor
To check if \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\) is a factor, we can set:

\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 = \frac{bc + ac + ab - abc}{abc} \]

For this to be a factor of \(D\), the expression \(bc + ac + ab - abc\) should equal zero when \(D\) is evaluated.

Since \(D = abc + ab + ac + bc\), we can see that \(bc + ac + ab - abc\) does not equal zero in general.

### Conclusion
Thus, we conclude that:

\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 \text{ is not a factor of } D. \]