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

Answer

To solve the problem, we start with the given determinant:

\[ \Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4 \]

We need to find the value of the determinant:

\[ D = \begin{vmatrix} a^3 - 1 & 0 & a - a^4 \\ 0 & a - a^4 & a^3 - 1 \\ a - a^4 & a^3 - 1 & 0 \end{vmatrix} \]

### Step 1: Factor out common terms
Notice that \(a - a^4 = a(1 - a^3)\). We can factor \(a(1 - a^3)\) out of the second and third rows.

\[ D = a(1 - a^3) \begin{vmatrix} a^3 - 1 & 0 & 1 \\ 0 & 1 & a^3 - 1 \\ 1 & a^3 - 1 & 0 \end{vmatrix} \]

### Step 2: Evaluate the new determinant
Now we will evaluate the determinant:

\[ D = a(1 - a^3) \begin{vmatrix} a^3 - 1 & 0 & 1 \\ 0 & 1 & a^3 - 1 \\ 1 & a^3 - 1 & 0 \end{vmatrix} \]

We can expand this determinant along the first row:

\[ = (a^3 - 1) \begin{vmatrix} 1 & a^3 - 1 \\ a^3 - 1 & 0 \end{vmatrix} - 0 + 1 \begin{vmatrix} 0 & 1 \\ 1 & a^3 - 1 \end{vmatrix} \]

Calculating the first determinant:

\[ \begin{vmatrix} 1 & a^3 - 1 \\ a^3 - 1 & 0 \end{vmatrix} = 1 \cdot 0 - (a^3 - 1)(a^3 - 1) = -(a^3 - 1)^2 \]

Calculating the second determinant:

\[ \begin{vmatrix} 0 & 1 \\ 1 & a^3 - 1 \end{vmatrix} = 0 \cdot (a^3 - 1) - 1 \cdot 1 = -1 \]

Putting it all together:

\[ D = (a^3 - 1)(-(a^3 - 1)^2) + 1(-1) = -(a^3 - 1)^3 - 1 \]

### Step 3: Substitute \(a^3 - 1\)
From the original determinant, we know that \(a^3 - 1 = \pm 4\). We need to evaluate \(D\) for both cases.

1. If \(a^3 - 1 = 4\):

\[ D = -4^3 - 1 = -64 - 1 = -65 \]

2. If \(a^3 - 1 = -4\):

\[ D = -(-4)^3 - 1 = 64 - 1 = 63 \]

### Final Answer
Thus, the value of the determinant \(D\) can be either \(-65\) or \(63\) depending on the value of \(a^3 - 1\).