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The given problem asks to prove that a determinant is zero given that \( x + y + z = 0 \). The matrix in question is:

\[
\text{Determinant of } \begin{vmatrix} 
1 & 1 & 1 \\ 
x & y & z \\ 
x^3 & y^3 & z^3 
\end{vmatrix} = 0
\]

### Proof:
We will compute the determinant of the given \( 3 \times 3 \) matrix and show that it is zero under the condition \( x + y + z = 0 \).

The determinant of a \( 3 \times 3 \) matrix is given by:

\[
\det = 1 \times \begin{vmatrix} y & z \\ y^3 & z^3 \end{vmatrix} 
- 1 \times \begin{vmatrix} x & z \\ x^3 & z^3 \end{vmatrix}
+ 1 \times \begin{vmatrix} x & y \\ x^3 & y^3 \end{vmatrix}
\]

#### Step 1: Compute each 2x2 minor
For the first minor:

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

For the second minor:

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

For the third minor:

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

#### Step 2: Substitute the minors back into the determinant expression
\[
\det = 1 \cdot yz(y^2 - z^2) - 1 \cdot xz(x^2 - z^2) + 1 \cdot xy(x^2 - y^2)
\]

Simplify the expression:

\[
\det = yz(y - z)(y + z) - xz(x - z)(x + z) + xy(x - y)(x + y)
\]

#### Step 3: Factor in \( x + y + z = 0 \)

Since \( x + y + z = 0 \), we know that \( x + y = -z \), \( y + z = -x \), and \( x + z = -y \). Substituting these into the determinant expression:

\[
\det = yz(y - z)(-x) - xz(x - z)(-y) + xy(x - y)(-z)
\]

Factor out the negative signs:

\[
\det = -xyz(y - z) + xyz(x - z) - xyz(x - y)
\]

Now, observe that each term involves a factor of \( xyz \), and we can factor that out:

\[
\det = xyz \left[ -(y - z) + (x - z) - (x - y) \right]
\]

Simplify the terms inside the brackets:

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

Thus, the determinant is:

\[
\det = xyz \times 0 = 0
\]

### Conclusion:
The determinant of the matrix is zero when \( x + y + z = 0 \), which completes the proof.

Would you like more details or have any questions?

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Here are 5 related questions for deeper understanding:

1. What happens if \( x + y + z \neq 0 \)? Can the determinant still be zero in specific cases?
2. How would this proof change if the entries in the matrix were powers other than \( 1, x, x^3 \)?
3. Can you generalize this determinant property for a \( n \times n \) matrix where the exponents form a specific pattern?
4. How does the Vandermonde determinant relate to this matrix form?
5. Can this approach be used to prove similar results for polynomials of degree higher than 3?

**Tip:** Always check for symmetry and conditions like \( x + y + z = 0 \) when simplifying determinants, as they often reduce the complexity of the computation.

If x + y + z = 0, then prove that the determinant of the following matrix is 0:

| 1   1   1 |
| x   y   z |
| x^3 y^3 z^3 |

This article provides a detailed proof showing that the determinant of a 3x3 matrix is zero given the condition x + y + z = 0. Learn the steps involved in computing the determinant and the mathematical concepts behind the solution.

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