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

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

Grade:9

1 Answers

Gursimran Singh
45 Points
5 years ago
if 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)))
 
 
 
 

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