If `|{:(1,x,x^(2)),(x,x^(2),1),(x^(2),1,x):}|` =3 then the value of `|{:(x^(3)-1,0,x-x^(4)),(0,x-x^(4),x^(3)-1),(x-x^(4),x^(3)-1,0):}|` is

Options

A. 6
B. 9
C. 18
D. 27

Answer

To solve the problem step by step, we will first analyze the determinants provided and then find the required value.

### Step 1: Analyze the first determinant

Given:
\[ D_1 = \begin{vmatrix} 1 & x & x^2 \\ x & x^2 & 1 \\ x^2 & 1 & x \end{vmatrix} = 3 \]

We need to compute this determinant.

### Step 2: Compute the first determinant

Using the determinant formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:
- \( a = 1, b = x, c = x^2 \)
- \( d = x, e = x^2, f = 1 \)
- \( g = x^2, h = 1, i = x \)

Calculating:
\[ D_1 = 1 \cdot (x^2 \cdot x - 1 \cdot 1) - x \cdot (x \cdot x - 1 \cdot x^2) + x^2 \cdot (x \cdot 1 - x^2 \cdot x) \]
\[ = 1 \cdot (x^3 - 1) - x \cdot (x^2 - x^2) + x^2 \cdot (x - x^3) \]
\[ = x^3 - 1 + x^2 \cdot (x - x^3) \]
\[ = x^3 - 1 + x^3 - x^5 \]
\[ = 2x^3 - x^5 - 1 \]

Setting this equal to 3:
\[ 2x^3 - x^5 - 1 = 3 \]
\[ -x^5 + 2x^3 - 4 = 0 \]

### Step 3: Analyze the second determinant

Now we need to compute:
\[ D_2 = \begin{vmatrix} x^3 - 1 & 0 & x - x^4 \\ 0 & x - x^4 & x^3 - 1 \\ x - x^4 & x^3 - 1 & 0 \end{vmatrix} \]

### Step 4: Compute the second determinant

Notice that we can factor out common terms. We can see that \(x - x^4 = x(1 - x^3)\).

Thus, we can factor out \(x(1 - x^3)\) from the first and third rows:
\[ D_2 = x(1 - x^3) \cdot \begin{vmatrix} x^3 - 1 & 0 & 1 \\ 0 & 1 & x^3 - 1 \\ 1 & x^3 - 1 & 0 \end{vmatrix} \]

### Step 5: Compute the new determinant

Now we compute the determinant:
\[ D = \begin{vmatrix} x^3 - 1 & 0 & 1 \\ 0 & 1 & x^3 - 1 \\ 1 & x^3 - 1 & 0 \end{vmatrix} \]

Using the determinant formula:
\[ D = (x^3 - 1) \begin{vmatrix} 1 & x^3 - 1 \\ x^3 - 1 & 0 \end{vmatrix} \]

Calculating the 2x2 determinant:
\[ = (x^3 - 1) \cdot (0 - (x^3 - 1)(1)) = (x^3 - 1)(-(x^3 - 1)) = -(x^3 - 1)^2 \]

### Step 6: Combine results

Thus:
\[ D_2 = x(1 - x^3)(-(x^3 - 1)^2) \]
\[ = -x(1 - x^3)(x^3 - 1)^2 \]

### Step 7: Substitute \(x^3 - 1\)

From our earlier result, we know that \(x^3 - 1 = \sqrt{3}\) (since \(D_1 = 3\)). Therefore:
\[ D_2 = -x(1 - x^3)(\sqrt{3})^2 \]
\[ = -x(1 - x^3) \cdot 3 \]

### Final Step: Evaluate

Since \(x^3 - 1 = 3\), we have \(x^3 = 4\), thus \(1 - x^3 = -3\):
\[ D_2 = -x(-3)(3) = 9x \]

### Conclusion

The value of the second determinant is:
\[ D_2 = 9 \]