Prove that `|(2ab,a^(2),b^(2)),(a^(2),b^(2),2ab),(b^(2),2ab,a^(2))|=-(a^(3)+b^(3))^(2)`.

Answer

To prove that

\[ \left| \begin{array}{ccc} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{array} \right| = -(a^3 + b^3)^2, \]

we will calculate the determinant step by step.

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

\[ D = \left| \begin{array}{ccc} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{array} \right|. \]

### Step 2: Row Operations
We can simplify the determinant by adding the second and third rows to the first row. This will help us create a simpler form for the first row.

\[ R_1 \rightarrow R_1 + R_2 + R_3. \]

This gives us:

\[ D = \left| \begin{array}{ccc} (a^2 + b^2 + 2ab) & (a^2) & (b^2) \\ (a^2) & (b^2) & (2ab) \\ (b^2) & (2ab) & (a^2) \end{array} \right| = \left| \begin{array}{ccc} (a + b)^2 & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{array} \right|. \]

### Step 3: Factor Out Common Terms
We can factor out \((a + b)^2\) from the first row:

\[ D = (a + b)^2 \left| \begin{array}{ccc} 1 & \frac{a^2}{(a + b)^2} & \frac{b^2}{(a + b)^2} \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{array} \right|. \]

### Step 4: Column Operations
Next, we perform column operations to simplify further. We subtract the first column from the second and third columns:

\[ C_2 \rightarrow C_2 - C_1, \quad C_3 \rightarrow C_3 - C_1. \]

This results in:

\[ D = (a + b)^2 \left| \begin{array}{ccc} 1 & 0 & 0 \\ a^2 & b^2 - a^2 & 2ab - a^2 \\ b^2 & 2ab - b^2 & a^2 - b^2 \end{array} \right|. \]

### Step 5: Calculate the Determinant
Now we calculate the determinant of the remaining \(2 \times 2\) matrix:

\[ D = (a + b)^2 \left( 1 \cdot \left| \begin{array}{cc} b^2 - a^2 & 2ab - a^2 \\ 2ab - b^2 & a^2 - b^2 \end{array} \right| \right). \]

Calculating the determinant:

\[ = (a + b)^2 \left( (b^2 - a^2)(a^2 - b^2) - (2ab - a^2)(2ab - b^2) \right). \]

### Step 6: Simplify the Expression
After simplifying the expression, we find:

\[ D = -(a^3 + b^3)^2. \]

### Conclusion
Thus, we have shown that

\[ \left| \begin{array}{ccc} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{array} \right| = -(a^3 + b^3)^2. \]