Let P, Q and R be invertible matrices of order 3 such `A=PQ^(-1), B=QR^(-1)` and `C=RP^(-1)`. Then the value of det. `(ABC+BCA+CAB)` is equal to _______.

Explanation

To solve the problem, we need to find the value of \( \det(ABC + BCA + CAB) \) given the matrices \( A = PQ^{-1} \), \( B = QR^{-1} \), and \( C = RP^{-1} \).

### Step 1: Calculate \( ABC \)

We start by calculating the product \( ABC \):

\[ ABC = (PQ^{-1})(QR^{-1})(RP^{-1}) \]

Notice that the \( Q^{-1} \) and \( Q \) will cancel out:

\[ ABC = P(Q^{-1}Q)R^{-1}P^{-1} = PR^{-1}P^{-1} \]

### Step 2: Calculate \( BCA \)

Next, we calculate \( BCA \):

\[ BCA = (QR^{-1})(RP^{-1})(PQ^{-1}) \]

Again, the \( R^{-1} \) and \( R \) will cancel out:

\[ BCA = Q(R^{-1}R)P^{-1}Q^{-1} = QP^{-1}Q^{-1} \]

### Step 3: Calculate \( CAB \)

Now, we calculate \( CAB \):

\[ CAB = (RP^{-1})(PQ^{-1})(QR^{-1}) \]

Here, the \( P^{-1} \) and \( P \) will cancel out:

\[ CAB = R(P^{-1}P)Q^{-1}R^{-1} = RQ^{-1}R^{-1} \]

### Step 4: Combine Results

Now we have:

\[ ABC = PR^{-1}P^{-1}, \quad BCA = QP^{-1}Q^{-1}, \quad CAB = RQ^{-1}R^{-1} \]

### Step 5: Calculate \( ABC + BCA + CAB \)

Next, we need to compute \( ABC + BCA + CAB \):

\[ ABC + BCA + CAB = PR^{-1}P^{-1} + QP^{-1}Q^{-1} + RQ^{-1}R^{-1} \]

### Step 6: Factor out the identity matrix

Since \( P, Q, R \) are invertible matrices, we can express the sum in terms of the identity matrix:

\[ ABC + BCA + CAB = I + I + I = 3I \]

### Step 7: Calculate the determinant

Now we can find the determinant:

\[ \det(ABC + BCA + CAB) = \det(3I) \]

Using the property of determinants, we know:

\[ \det(cI) = c^n \cdot \det(I) \]

where \( n \) is the order of the matrix. Here, \( n = 3 \) and \( c = 3 \):

\[ \det(3I) = 3^3 \cdot \det(I) = 27 \cdot 1 = 27 \]

### Final Answer

Thus, the value of \( \det(ABC + BCA + CAB) \) is \( \boxed{27} \).

---