Question

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

Answer 1

 a+b+c=0 the (a+b+c)^2 = 0

a^2+b^2+c^2+2(ab+bc+ca) = 0

a^2+b^2+c^2 = -2(ab+bc+ca) squaring on both sides

a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+ca^2b))

a^4+b^4+c^4 = 4(a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+ca^2b))-2(a^2b^2+b^2c^2+c^2a^2)

= 2[2(a^2b^2+b^2c^2+c^2a^2+(ab^2c+bc^2a+ca^2b))-(a^2b^2+b^2c^2+c^2a^2)]

= 2((a^2b^2+b^2c^2+c^2a^2+(ab^2c+bc^2a+ca^2b)))