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:   $\boxed{(p, q) = (7,\ 3)}$