Question

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

Answer 1

This potential solution comes courtesy of @Ian
Please check it and determine if you agree.

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

(a^2 + b^2 + c^2)^2

= [(-(b+c))^2 + b^2 + c^2]^2

= [(b+c)^2 + b^2 + c^2]^2

= (b^2 + 2bc + c^2 + b^2 + c^2)^2

= (2b^2 + 2bc + 2c^2)^2

= (2b^2 + 2bc + 2c^2)^2

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

= 4(b^4 + b^2c^2 + c^4 + 2b^3c + 2b^2c^2 + 2bc^3)

= 4(b^4 + c^4 + 2b^3c + 3b^2c^2 + 2bc^3)

= 2(2b^4 + 2c^4 + 4b^3c + 6b^2c^2 + 4bc^3)

= 2(b^4 + 4b^3c + 6b^2c^2 + 4bc^3 + c^4 + b^4 + c^4)

= 2[(b+c)^4 + b^4 + c^4]

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

= 2(a^4 + b^4 + c^4).

Answer 2

Read the solution in context at this link: http://tinyurl.com/c5rv7nz

Answer 3

a+b+c=0 square it ... put the ab + .. terms on one side, then square it .. you might get it.

Answer 4

plse explain last 3 rd step

Answer 5

This one?

Answer 6

Last three steps that is
= 2[(b+c)^4 + b^4 + c^4]
= 2[(-(b+c))^4 + b^4 + c^4]
= 2(a^4 + b^4 + c^4).

Answer 7

I did straightforward:\[(a+b+c)^2=a^2+ab+ac+ab+b^2+bc+ac+bc+c^2\]\[=a^2+b^2+c^2+2(ab+bc+ac)=0\]\[a^2+b^2+c^2=-2(ab+bc+ac)\]Now square again. Left hand side:\[(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)\]Right hand side:\[4(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2)\]So:\[a^4+b^4+c^4=-2(a^2b^2+b^2c^2+a^2c^2)+4(a^2b^2+b^2c^2+a^2c^2)+8abc(a+b+c)\]Or:\[a^4+b^4+c^4=2(a^2b^2+b^2c^2+a^2c^2)+8abc(a+b+c)\]

Answer 8

Because a+b+c=0, we have:\[a^4+b^4+c^4=2(a^2b^2+b^2c^2+a^2c^2)\]