Maximum Value of \( f(x) = (x - 1)^2(x + 1)^3 \)

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let \( u = (x - 1)^2 \), \( v = (x + 1)^3 \)
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set \( f'(x) = 0 \): \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate \( f(x) \) at these points

• \( f(1) = 0 \)
• \( f(-1) = 0 \)
• \( f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 \)

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is \( \frac{3456}{3125} = 2^p \cdot 3^q / 3125 \)

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer:   \( \boxed{(p, q) = (7,\ 3)} \)