The value of $[(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]÷[(a-b)^3+(b-c)^3+(c-a)^3]$ is equal to :

(Given a ≠ b ≠ c)

$(a+b)(b+c)(c+a)$

The value of $[(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]÷[(a-b)^3+(b-c)^3+(c-a)^3]$ is equal to

We know when  a³ + b³ + c³  is equal to 0 then the value of is a³ + b³ + c³  = 3abc

and we also know that,

a² - b² = (a + b) (a – b)

So according to the question,

$[(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]÷[(a-b)^3+(b-c)^3+(c-a)^3]$ = $3[(a^2-b^2)+(b^2-c^2)+(c^2-a^2)]÷3[(a-b)+(b-c)+(c-a)]$

= $3[(a-b)(a+b)+(b+c)(b-c)+(c+a)(c-a)]÷3[(a-b)+(b-c)+(c-a)]$

= $3[(a-b)(a+b)+(b+c)(b-c)+(c+a)(c-a)]÷3[(a-b)+(b-c)+(c-a)]$

= $(a+b)(b+c)(c+a)$