let `|{:(1+x,x,x^(2)),(x,1+x,x^(2)),(x^(2),x,1+x):}|=(1)/(6)(x-alpha_(1))(x-alpha_(2))(x-alpha_(3))(x-alpha_(4))` be an identity in x, where `alpha_(1),alpha_(2),alpha_(3),alpha_(4)` are independent of x. Then find the value of `alpha_(1)alpha_(2)alpha_(3)alpha_(4)`

Explanation

To solve the given problem, we need to evaluate the determinant and relate it to the expression on the right-hand side of the equation. The steps are as follows:

### Step 1: Write the Determinant
We have the determinant:
\[ D = \begin{vmatrix} 1+x & x & x^2 \\ x & 1+x & x^2 \\ x^2 & x & 1+x \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand this determinant using the method of cofactor expansion. However, for simplicity, we can directly compute it using properties of determinants or row operations.

### Step 3: Substitute \( x = 0 \)
To simplify the calculation, we substitute \( x = 0 \) into the determinant:
\[ D(0) = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \]
Calculating this determinant gives us:
\[ D(0) = 1 \]

### Step 4: Substitute \( x = 0 \) in the Right-Hand Side
Now, substituting \( x = 0 \) into the right-hand side:
\[ \frac{1}{6} (0 - \alpha_1)(0 - \alpha_2)(0 - \alpha_3)(0 - \alpha_4) = \frac{1}{6} (-\alpha_1)(-\alpha_2)(-\alpha_3)(-\alpha_4) = \frac{1}{6} \alpha_1 \alpha_2 \alpha_3 \alpha_4 \]

### Step 5: Set the Determinants Equal
Since the original equation is an identity, we set the left-hand side equal to the right-hand side:
\[ 1 = \frac{1}{6} \alpha_1 \alpha_2 \alpha_3 \alpha_4 \]

### Step 6: Solve for \( \alpha_1 \alpha_2 \alpha_3 \alpha_4 \)
Multiplying both sides by 6 gives:
\[ \alpha_1 \alpha_2 \alpha_3 \alpha_4 = 6 \]

### Final Answer
Thus, the value of \( \alpha_1 \alpha_2 \alpha_3 \alpha_4 \) is:
\[ \boxed{6} \]