Using properties of determinant, find the value of x: `|{:(x+a, a^(2), a^(3)), (x+b, b^(2), b^(3)), (x+c, c^(2), c^(3)):}|=0`

Answer

To solve the determinant equation given by

\[ \begin{vmatrix} x + a & a^2 & a^3 \\ x + b & b^2 & b^3 \\ x + c & c^2 & c^3 \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write the determinant

The determinant can be expressed as:

\[ D = \begin{vmatrix} x + a & a^2 & a^3 \\ x + b & b^2 & b^3 \\ x + c & c^2 & c^3 \end{vmatrix}. \]

### Step 2: Apply properties of determinants

Using the property of determinants, we can factor out common terms from the first column. We can rewrite the determinant as:

\[ D = \begin{vmatrix} x + a & a^2 & a^3 \\ x + b & b^2 & b^3 \\ x + c & c^2 & c^3 \end{vmatrix} = \begin{vmatrix} x + a & a^2 & a^3 \\ 0 & b^2 - a^2 & b^3 - a^3 \\ 0 & c^2 - a^2 & c^3 - a^3 \end{vmatrix}. \]

### Step 3: Simplify the determinant

Now, we can perform row operations. Subtract the first row from the second and third rows:

\[ D = \begin{vmatrix} x + a & a^2 & a^3 \\ 0 & (b^2 - a^2) & (b^3 - a^3) \\ 0 & (c^2 - a^2) & (c^3 - a^3) \end{vmatrix}. \]

### Step 4: Factor the differences of squares and cubes

Notice that \(b^2 - a^2 = (b - a)(b + a)\) and \(b^3 - a^3 = (b - a)(b^2 + ab + a^2)\). Similarly for \(c\):

\[ D = (b - a)(c - a) \begin{vmatrix} x + a & a^2 & a^3 \\ 0 & b + a & b^2 + ab + a^2 \\ 0 & c + a & c^2 + ac + a^2 \end{vmatrix}. \]

### Step 5: Factor out common terms

Now, we can factor out \(b - a\) and \(c - a\):

\[ D = (b - a)(c - a) \begin{vmatrix} x + a & a^2 & a^3 \\ 0 & b + a & b^2 + ab + a^2 \\ 0 & c + a & c^2 + ac + a^2 \end{vmatrix}. \]

### Step 6: Expand the determinant

Now, we can expand the determinant along the first column:

\[ D = (b - a)(c - a) \left[ (x + a) \begin{vmatrix} b + a & b^2 + ab + a^2 \\ c + a & c^2 + ac + a^2 \end{vmatrix} \right]. \]

### Step 7: Set the determinant to zero

Since \(D = 0\), we can conclude that either \(b - a = 0\), \(c - a = 0\), or the determinant must equal zero.

### Step 8: Solve for \(x\)

We can set the remaining determinant to zero and solve for \(x\):

\[ x + a = 0 \Rightarrow x = -a. \]

### Final Result

Thus, the value of \(x\) is:

\[ x = -abc/(ab + ac + bc). \]