Prove that if a + b + c = 0 then (a^2 + b^2 + c^2)^2 = 2(a^4 + b^4 + c^4)

We have
(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2a^2b^2 + 2 b^2 c^2 + 2 c^2 a^2 ... 1

Now a^2 b^2 + a^2 c^2 = a^2((b+c)^2 – 2bc) = a^4 – 2a^2bc .2 (as b+c = - a)

Similarly
b^2 c^2 + b^2 a^2 = b^4 – 2b^2ac ..3
c^2a^2 + c^2b^2 = c^4 – 2c^abc ...4

from (2) (3) and (4)
2(a^2b^2 + b^2 c^2 + c^2 a^2) = (a^4 + b^4 + c^4 – 2abc(a+b+c))
= a^4 + b^4 + c^4 ...5
as a+b+c = 0
Putting value of  2(a^2b^2 + b^2 c^2 + c^2 a^2) from (5) in (1) we get the result