If `x!=y!=z` and `|xx^2 1+x^3y y^2 1+y^3z z^2 1+z^3|=0` , then prove that `x y z=-1` .

Explanation

To solve the problem, we need to prove that if the determinant

\[ D = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ 1+x^3 & 1+y^3 & 1+z^3 \end{vmatrix} = 0 \]

then \(xyz = -1\).

### Step 1: Write the Determinant
We start with the determinant:

\[ D = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ 1+x^3 & 1+y^3 & 1+z^3 \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can separate the last row into two parts:

\[ D = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} + \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Notice that we can factor out \(xyz\) from the first determinant:

\[ D = xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} + \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} \]

### Step 4: Use Properties of Determinants
The determinant of the second matrix can be expressed as:

\[ \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = (x-y)(y-z)(z-x) \]

### Step 5: Combine the Results
So we have:

\[ D = xyz \cdot 0 + (x-y)(y-z)(z-x) = 0 \]

### Step 6: Analyze the Conditions
Since \(x \neq y \neq z\), the term \((x-y)(y-z)(z-x) \neq 0\). Therefore, for the determinant \(D\) to be zero, we must have:

\[ 1 + xyz = 0 \]

### Step 7: Conclude the Proof
This implies:

\[ xyz = -1 \]

Thus, we have proved that if the determinant is zero, then \(xyz = -1\).

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