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