Using properties of determinants , find the value of `k` if
`|{:(x,y,x+y),(y,x+y,x),(x+y,x,y):}|=k(x^(3)+y^(3))`.

Answer

To find the value of \( k \) in the equation

\[ \left| \begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array} \right| = k(x^3 + y^3), \]

we will use properties of determinants.

### Step 1: Rewrite the determinant
We start with the determinant:

\[ D = \left| \begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array} \right| \]

We can simplify the determinant by manipulating the columns. We will replace the first column with the sum of all three columns:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D = \left| \begin{array}{ccc} x + y + (x+y) & y & x \\ y + (x+y) + x & x+y & x \\ (x+y) + x + y & x & y \end{array} \right| \]

This simplifies to:

\[ D = \left| \begin{array}{ccc} 2(x+y) & y & x \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{array} \right| \]

### Step 2: Factor out common terms
Notice that \( 2(x+y) \) is common in the first column. We can factor it out:

\[ D = 2(x+y) \left| \begin{array}{ccc} 1 & y & x \\ 1 & x+y & x \\ 1 & x & y \end{array} \right| \]

### Step 3: Simplify the determinant
Now we will simplify the determinant:

\[ \left| \begin{array}{ccc} 1 & y & x \\ 1 & x+y & x \\ 1 & x & y \end{array} \right| \]

We can perform row operations. Subtract the first row from the second and third rows:

\[ = \left| \begin{array}{ccc} 1 & y & x \\ 0 & (x+y - y) & (x - x) \\ 0 & (x - y) & (y - x) \end{array} \right| \]

This simplifies to:

\[ = \left| \begin{array}{ccc} 1 & y & x \\ 0 & x & 0 \\ 0 & (x - y) & (y - x) \end{array} \right| \]

### Step 4: Calculate the determinant
Now we can calculate the determinant:

\[ = 1 \cdot \left| \begin{array}{cc} x & 0 \\ x - y & y - x \end{array} \right| \]

Calculating this 2x2 determinant:

\[ = 1 \cdot (x(y - x) - 0) = xy - x^2 \]

### Step 5: Substitute back into the determinant
Now substituting back, we have:

\[ D = 2(x+y)(xy - x^2) \]

### Step 6: Use the identity for \( x^3 + y^3 \)
Recall that:

\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]

We can express \( xy - x^2 \) in terms of \( x^3 + y^3 \):

\[ D = 2(x+y)(-1)(x^2 - xy + y^2) = -2(x+y)(x^2 - xy + y^2) \]

### Step 7: Equate to find \( k \)
Now, we compare this with \( k(x^3 + y^3) \):

\[ -2(x+y)(x^2 - xy + y^2) = k(x^3 + y^3) \]

### Step 8: Solve for \( k \)
From the identity \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \), we can conclude:

\[ k = -2 \]

Thus, the value of \( k \) is

\[ \boxed{-2}. \]