Prove that: (a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a) = 2(a³ + b³ + c³ - 3abc)

We know that –

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

So applying the theorem here,

(a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a) = ((a + b) + (b + c) + (c + a))((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c) - (b + c)(c + a)-(c + a)(a + b))

= (2(a + b + c))((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c)-(b + c)(c + a)-(c + a)(a + b))

{since ((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c) - (b + c)(c + a) - (c + a)(a + b))

= (a² + b² + c² – ab – bc - ca)}

= 2 (a³ + b³ + c³ - 3(a)(b)(c))

{using this theorem again: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)}