Question 53521
This problem is easy to get lost doing, so I broke it up into smaller problems and  then grouped all my like terms vertically in a table.  Unfortunately, I can't put tables into my responses.  Hopefully, I can be organized enough that you don't get lost.  
First, substitute (a+b) everywhere that you have a c in the problem.
{{{a^4+b^4+c^4-2b^2c^2-2c^2a^2-2a^2b^2=0}}}
{{{a^4+b^4+(a+b)^4-2b^2(a+b)^2-2(a+b)^2a^2-2a^2b^2=0}}}
This is where I broke the problem up into smaller problems:
{{{(a+b)^2=(a+b)(a+b)}}}
{{{(a+b)^2=a^2+ab+ab+b^2}}}
{{{(a+b)^2=a^2+2ab+b^2}}}
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If you don't know the formula for this you can just distribute:
{{{(a+b)^4=(a+b)^2(a+b)^2}}}
{{{(a+b)^4=(a^2+2ab+b^2)(a^2+2ab+b^2)}}}
{{{(a+b)^4=a^2(a^2)+a^2(2ab)+a^2(b^2)+2ab(a^2)+2ab(2ab)+2ab(b^2)+b^2(a^2)+b^2(2ab)+b^2(b^2)}}}
{{{(a+b)^4=a^4+2a^3b+a^2b^2+2a^3b+4a^2b^2+2ab^3+a^2b^2+2ab^3+b^4}}}
{{{(a+b)^4=a^4+(2+2)a^3b+(1+4+1)a^2b^2+(2+2)ab^3+b^4}}}
{{{(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4}}}
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Substitute those back into the problem and you have:
{{{a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)-2b^2(a^2+2ab+b^2)-2(a^2+2ab+b^2)a^2-2a^2b^2=0}}}
{{{a^4+b^4+a^4+4a^3b+6a^2b^2+4ab^3+b^4-2a^2b^2-4ab^3-2b^4-2a^4-4a^3b-2a^2b^2-2a^2b^2=0}}}
{{{(1+1-2)a^4+(1+1-2)b^4+(4-4)a^3b+(6-2-2-2)a^2b^2+(4-4)ab^3=0}}}
{{{0a^4+0b^4+0a^3b+0a^2b^2+0ab^3=0}}}
0=0
I hope I caught all of my type-0's.  Happy studying!