Let us factorise

Let us consider (3a - 2b) = x, (2b - 5c) = y, (5c - 3a) = z

So x + y + z = 3a – 2b + 2b – 5c + 5c – 3a = 0

(3a – 2b) ³ + (2b – 5c) ³ + (5c – 3a) ³ = x³ + y³ + z³

Since x + y + z = 0, hence

x³ + y³ + z³ = 3xyz

⇒ (3a – 2b) ³ + (2b – 5c) ³ + (5c – 3a) ³ = 3×(3a – 2b)×(2b – 5c)× (5c – 3a)

3(3a – 2b) (2b – 5c) (5c – 3a)