Factorize (x – 2y)³ + (2y – 3z)³ + (3z – x)³ without finding the cubes explicitly.

As the expression is of the form of sum of three cubes, for factorizing, we will use the identity

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

Taking the 3abc term to the right hand side, we have

a³ + b³ + c³ = (a + b + c)(a² + b² + c² – ab – bc – ca) + 3abc

Here, a = (x – 2y), b = (2y – 3z) & c = (3z – x)

⇒ a³ + b³ + c³ = (x – 2y)³ + (2y – 3z)³ + (3z – x)³

Observe that a + b + c = (x – 2y) + (2y – 3z) + (3z – x) = 0

⇒ a³ + b³ + c³ = 0 + 3abc = 3abc

∴ (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x)