If HCF of 768 and x⁶y² is 32xy for natural numbers x ≥ 2, y ≥ 2, then what is the value of (x + y) ?

768 = 2⁸ × 3. HCF(768, x⁶y²) = 32xy = 2⁵ × xy. For HCF to equal 2⁵xy, we need x⁶y² to contain 2⁵xy as factor and HCF with 2⁸·3 to be 2⁵xy. So xy must be a power of 2 or include 3. Since HCF must divide 768 = 2⁸×3, 32xy must divide 768, so xy divides 24. With x,y ≥ 2: possible (x,y): testing x=2, y=3: x⁶y² = 64·9 = 576. HCF(768, 576) = 192. 32xy = 192. ✓. So x+y = 5.