Question

[a2-b2] [a2+b2]+ [b2-c2] [b2-c2]+ [c2-a2] [c2-a2=0

Answer 1

Explanation:

we know that (x+y)(x-y)=x^2-y^2

(a^2-b^2)(a^2+b^2) = a^4 - b^4

(b^2-c^2)(b^2+c^2) = b^4 - c^4

(c^2-a^2)(c^2+a^2) = c^4 - a^4

so now we have to add all these

= a^4-b^4+b^4-c^4+c^4-a^4

= a^4-a^4+b^4-b^4+c^4-c^4

= 0

Hence proved