If `|(x^(n),x^(n+2),x^(n+3)),(y^(n),y^(n+2),y^(n+3)),(z^(n),z^(n+2),z^(n+3))|`
`= (x -y) (y -z) (z -x) ((1)/(x) + (1)/(y) + (1)/(z))`, then n equals

Options

A. 1
B. `-1`
C. 2
D. `-2`

Explanation

To solve the given determinant equation, we need to find the value of \( n \) such that:

\[ |(x^n, x^{n+2}, x^{n+3}), (y^n, y^{n+2}, y^{n+3}), (z^n, z^{n+2}, z^{n+3})| = (x - y)(y - z)(z - x) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \]

**Step 1: Calculate the determinant.**

The determinant can be expanded as follows:

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

Using properties of determinants, we can factor out common terms from each column:

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

**Step 2: Calculate the simplified determinant.**

Now, we compute the determinant of the simplified matrix:

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

Using the determinant formula for a \( 3 \times 3 \) matrix, we have:

\[ D' = 1 \cdot (y^2 z^3 - y^3 z^2) - 1 \cdot (x^2 z^3 - x^3 z^2) + 1 \cdot (x^2 y^3 - x^3 y^2) \]

This simplifies to:

\[ D' = (y^2 z^3 - y^3 z^2) - (x^2 z^3 - x^3 z^2) + (x^2 y^3 - x^3 y^2) \]

Factoring out common terms, we can express it as:

\[ D' = (y - z)(z - x)(x - y) \]

**Step 3: Combine results.**

Thus, we can write:

\[ D = (y - z)(z - x)(x - y) \cdot x^n y^n z^n \]

**Step 4: Set the equation equal to the given expression.**

Now we equate this to the right-hand side of the original equation:

\[ (y - z)(z - x)(x - y) \cdot x^n y^n z^n = (x - y)(y - z)(z - x) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \]

**Step 5: Simplify and solve for \( n \).**

Notice that both sides have the factor \( (y - z)(z - x)(x - y) \). We can cancel this out (assuming \( x, y, z \) are distinct):

\[ x^n y^n z^n = \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \]

The right-hand side can be rewritten as:

\[ \frac{yz + zx + xy}{xyz} \]

Thus, we have:

\[ x^n y^n z^n = \frac{yz + zx + xy}{xyz} \]

Multiplying both sides by \( xyz \):

\[ x^{n+1} y^{n+1} z^{n+1} = yz + zx + xy \]

**Step 6: Determine the value of \( n \).**

To satisfy this equation for all \( x, y, z \), we can see that \( n + 1 \) must equal 0, leading to:

\[ n + 1 = 0 \implies n = -1 \]

Thus, the final answer is:

\[ \boxed{-1} \]