Question

If a + b + c = 0 then prove that a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + c²a²)

Answer 1

a+b+c=0
squaring both sides,
(a+b)²+2(a+b)c+c²=0
or, a²+2ab+b²+2ac+2bc+c²=0
or, a²+b²+c²=-2(ab+bc+ac) -------------------(1)
∴, a⁴+b⁴+c⁴
=(a²)²+(b²)²+(c²)²
={(a²+b²)²-2a²b²}+(c²)²
=(a²+b²)²+(c²)²-2a²b²
={(a²+b²+c²)²-2(a²+b²)c²}-2a²b²
={-2(ab+bc+ac)}²-2a²c²-2b²c²-2a²b² [using (1)]
=4[(ab+bc)²+2(ab+bc)ac+a²c²]-2a²c²-2b²c²-2a²b²
=4(a²b²+2.ab.bc+b²c²+2a²bc+2abc²+a²c²)-2a²c²-2b²c²-2a²b²
=4a²b²+4b²c²+4a²c²-2a²c²-2b²c²-2a²b²+8a²bc+8ab²c+8abc²
=2a²b²+2b²c²+2a²c²+8abc(a+b+c)
=2(a²b²+b²c²+c²a²) (Proved)
[∵, a+b+c=0]