Answer: Both I and II

Set up using HCF × LCM = product of the two numbers.

Let the numbers be $a$ and $b$ (distinct). We know: $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \cdot b$

Given this product equals the cube of one of the numbers, say $a^3$: $a \cdot b = a^3 \Rightarrow b = a^2$

So the two numbers are $a$ and $a^2$ (with $a \ge 2$ for them to be distinct positive integers).

Check the statements.

• I. Difference is even. $a^2 - a = a(a-1)$, the product of two consecutive integers, which is always even. ✓ True
• II. One of the numbers is a perfect square. $b = a^2$ is a perfect square. ✓ True

Sanity check: $a=2 \Rightarrow$ numbers $2, 4$; HCF $=2$, LCM $=4$, product $=8 = 2^3$. Difference $=2$ (even), and $4$ is a perfect square. ✓