If `|(x^k,x^(k+2),x^(k+3)), (y^k,y^(k+2),y^(k+3)), (z^k,z^(k+2),z^(k+3))|=(x-y)(y-z)(z-x){1/x+1/y+1/z}` then `k=`

Answer

To solve the problem, we need to evaluate the determinant and simplify it according to the given condition. Let's break it down step by step.

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} \]

### Step 2: Factor Out Common Terms
We can factor out \(x^k\), \(y^k\), and \(z^k\) from each row:
\[ D = x^k y^k z^k \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix} \]

### Step 3: Simplify the Determinant
Now we simplify the determinant:
\[ D = x^k y^k z^k \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix} \]

We can perform row operations to simplify this determinant. Subtract the first row from the second and third rows:
\[ D = x^k y^k z^k \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & y^2 - x^2 & y^3 - x^3 \\ 0 & z^2 - x^2 & z^3 - x^3 \end{vmatrix} \]

### Step 4: Factor Differences of Powers
Using the identities \(a^2 - b^2 = (a-b)(a+b)\) and \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\):
\[ D = x^k y^k z^k \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & (y-x)(y+x) & (y-x)(y^2 + xy + x^2) \\ 0 & (z-x)(z+x) & (z-x)(z^2 + zx + x^2) \end{vmatrix} \]

### Step 5: Factor Out Common Terms
Now we can factor out \((y-x)\) and \((z-x)\):
\[ D = x^k y^k z^k (y-x)(z-x) \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & y+x & y^2 + xy + x^2 \\ 0 & z+x & z^2 + zx + x^2 \end{vmatrix} \]

### Step 6: Evaluate the Remaining Determinant
The remaining determinant can be evaluated:
\[ D = x^k y^k z^k (y-x)(z-x) \cdot \text{(some expression)} \]

### Step 7: Compare with Given Expression
According to the problem, we have:
\[ D = (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \]

### Step 8: Equate and Solve for \(k\)
Comparing both sides, we find:
\[ xyz^{k} \text{(some expression)} = (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \]

From the comparison, we can deduce:
\[ k = -1 \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{-1} \]