Let `A=[[2,0,7] , [0,1,0], [1,-2,1]]` and `B=[[-x,14x,7x] , [0,1,0] , [x,-4x,-2x]]` are two matrices such that `AB=(AB)^(-1)` and `AB!=I` then `Tr((AB)+(AB)^2+(AB)^3+(AB)^4+(AB)^5+(AB)^6)=`

Explanation

To solve the problem, we need to find the trace of the matrix expression \( Tr(AB + AB^2 + AB^3 + AB^4 + AB^5 + AB^6) \) given that \( AB = (AB)^{-1} \) and \( AB \neq I \).

### Step 1: Calculate the product \( AB \)

Given matrices:
\[ A = \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} \]

To find \( AB \), we multiply matrix \( A \) by matrix \( B \):

\[ AB = \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} \]

Calculating each element of the resulting matrix:

1. First row:
- First column: \( 2(-x) + 0 + 7x = -2x + 7x = 5x \)
- Second column: \( 2(14x) + 0 + 7(-4x) = 28x - 28x = 0 \)
- Third column: \( 2(7x) + 0 + 7(-2x) = 14x - 14x = 0 \)

2. Second row:
- First column: \( 0(-x) + 1(0) + 0(x) = 0 \)
- Second column: \( 0(14x) + 1(1) + 0(-4x) = 1 \)
- Third column: \( 0(7x) + 1(0) + 0(-2x) = 0 \)

3. Third row:
- First column: \( 1(-x) + (-2)(0) + 1(x) = -x + x = 0 \)
- Second column: \( 1(14x) + (-2)(1) + 1(-4x) = 14x - 2 - 4x = 10x - 2 \)
- Third column: \( 1(7x) + (-2)(0) + 1(-2x) = 7x - 2x = 5x \)

Thus, we have:
\[ AB = \begin{bmatrix} 5x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10x - 2 & 5x \end{bmatrix} \]

### Step 2: Use the property \( AB = (AB)^{-1} \)

Since \( AB = (AB)^{-1} \), we can conclude that \( (AB)(AB) = I \):
\[ (AB)^2 = I \]

### Step 3: Calculate the powers of \( AB \)

From \( (AB)^2 = I \), we can deduce:
- \( (AB)^3 = AB \)
- \( (AB)^4 = I \)
- \( (AB)^5 = AB \)
- \( (AB)^6 = I \)

### Step 4: Substitute into the trace expression

Now substituting back into the trace expression:
\[ Tr(AB + AB^2 + AB^3 + AB^4 + AB^5 + AB^6) = Tr(AB + I + AB + I + AB + I) \]
This simplifies to:
\[ Tr(3AB + 3I) = 3Tr(AB) + 3Tr(I) \]
Since \( Tr(I) = 3 \) (the trace of a 3x3 identity matrix), we have:
\[ Tr(AB + AB^2 + AB^3 + AB^4 + AB^5 + AB^6) = 3Tr(AB) + 9 \]

### Step 5: Calculate \( Tr(AB) \)

The trace of \( AB \) is the sum of its diagonal elements:
\[ Tr(AB) = 5x + 1 + (10x - 2) = 15x - 1 \]

### Step 6: Final expression for the trace

Substituting \( Tr(AB) \) back into the expression:
\[ Tr(AB + AB^2 + AB^3 + AB^4 + AB^5 + AB^6) = 3(15x - 1) + 9 = 45x - 3 + 9 = 45x + 6 \]

### Final Answer
Thus, the final result is:
\[ \boxed{45x + 6} \]