Simplify: .

To simplify this expression, let us evaluate the numerator and denominator separately.

First, consider numerator (a² – b²)³ + (b² – c²)³ + (c² – a²)³

As the expression is of the form of sum of three cubes, for

factorizing, we will use the identity

x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

Taking the 3xyz term to the right-hand side, we have

x³ + y³ + z³ = (x + y + z)(x² + y² + z² – xy – yz – zx) + 3xyz

Here, x = (a² – b²), y = (b² – c²) & z = (c² – a²)

⇒ x³ + y³ + z³ = (a² – b²)³ + (b² – c²)³ + (c² – a²)³

Observe that x + y + z = (a² – b²) + (b² – c²) + (c² – a²) = 0

⇒ x³ + y³ + z³ = 0 + 3xyz = 3xyz

∴ (a² – b²)³ + (b² – c²)³ + (c² – a²)³ = 3(a² – b²)(b² – c²)(c² – a²)

Now, consider the denominator (a – b)³ + (b – c)³ + (c – a)³

Once again, we use the same identity as above, that is

x³ + y³ + z³ = (x + y + z)(x² + y² + z² – xy – yz – zx) + 3xyz

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

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

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

⇒ x³ + y³ + z³ = 0 + 3xyz = 3xyz

∴ (a – b)³ + (b – c)³ + (c – a)³ = 3(a – b)(b – c)(c – a)

So, using both the numerator and denominator we found,

But, we have,

a² – b² = (a + b)(a – b)

b² – c² = (b + c)(b – c)

c² – a² = (c + a)(c – a).