Question

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0​

Answer 1

Step-by-step explanation:

a3 + b3 + c3 - 3abc 

= (a2 + b2 + c2 - ab - bc - ca)(a + b + c) 

= a2 + b2 + c2

= ab + bc + ca (given)

⇒ (a2 + b2 + c2 - ab - bc - ca) = 0

∴ a3 + b3 + c3 - 3abc 

(a + b + c) x 0 = 0

∴ a3 + b3 + c3 = 3abc

Answer 2

Answer:

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0.

(a²-b²)+(b²-c²)+(c²-a²)=0. (Because (a-b)(a+b)=a²-b²).

a²-b²+b²-c²+c²-a²=0. (Because +×-=-).

a²-a²+b²-b²+c²-c²=0.

a²-a² will cancel as well as b²-b² will also cancel as well as c²-c² will also be cancelled. (Because +×-=-).

So, 0=0.

I think this is your answer.