n = ab ab=a2+b3 ⇒10a+b=a2+b3 ⇒ a (10 - a) + b (1 - b) (1 + b) = 0 ⇒ a (10 – a) = (b – 1) (b) (b + 1) If b = 2; a (10 – a) = 6 ⇒ no value of 'a' b = 3; a (10 – a) = 24 ⇒ a ∈ {4, 6}. {nos. are 43 & 63} b = 4; a (10 – a) = 60 ⇒ no value of a b = 5; a (10 – a) = 120 ⇒ no need to check further ∴ nos. are 43 & 63.