Suppose p,q,r `in` R and and pqr = 2.5 . Let
A = `[(p,q,r),(r,p,q),(q,r,p)]`
If `"AA"= I_(3)` then maximum possible value of `p^(3)+q^(3)+r^(3)` is

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript while ensuring clarity and completeness.

### Step 1: Understand the Given Information
We are given that \( p, q, r \in \mathbb{R} \) and \( pqr = 2.5 \). We also have the matrix:
\[ A = \begin{pmatrix} p & q & r \\ r & p & q \\ q & r & p \end{pmatrix} \]
and it is given that \( AA = I_3 \), where \( I_3 \) is the identity matrix of order 3.

### Step 2: Find the Determinant of Matrix A
Since \( AA = I \), we can take the determinant of both sides:
\[ \det(AA) = \det(I) \implies \det(A)^2 = 1 \]
This implies that:
\[ \det(A) = 1 \quad \text{or} \quad \det(A) = -1 \]

### Step 3: Calculate the Determinant of A
To find \( \det(A) \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix. The determinant of matrix \( A \) is given by:
\[ \det(A) = p \begin{vmatrix} p & q \\ r & p \end{vmatrix} - q \begin{vmatrix} r & q \\ q & p \end{vmatrix} + r \begin{vmatrix} r & p \\ q & r \end{vmatrix} \]
Calculating the minors:
\[ = p(p^2 - qr) - q(rp - q^2) + r(r^2 - pq) \]
This simplifies to:
\[ = p^3 + q^3 + r^3 - 3pqr \]

### Step 4: Set Up the Equations for Determinant
From the determinant calculation, we have:
\[ \det(A) = p^3 + q^3 + r^3 - 3pqr \]
We can set this equal to both possible values of the determinant:
1. If \( \det(A) = 1 \):
\[ p^3 + q^3 + r^3 - 3(2.5) = 1 \implies p^3 + q^3 + r^3 = 1 + 7.5 = 8.5 \]
2. If \( \det(A) = -1 \):
\[ p^3 + q^3 + r^3 - 3(2.5) = -1 \implies p^3 + q^3 + r^3 = -1 + 7.5 = 6.5 \]

### Step 5: Determine the Maximum Value
Now we compare the two results:
- From \( \det(A) = 1 \), we found \( p^3 + q^3 + r^3 = 8.5 \).
- From \( \det(A) = -1 \), we found \( p^3 + q^3 + r^3 = 6.5 \).

Thus, the maximum possible value of \( p^3 + q^3 + r^3 \) is:
\[ \boxed{8.5} \]