Question

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

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Answer 1

A+B+C=0
A+B=–C
squaring on both sides
(A+B)² = (–C)²
A²+B²+2AB = C²
A²+B²–C²=–2AB
Again squaring on both sides
(A²+B²–C²)²=(–2AB)²
A⁴+B⁴+C⁴+2A²B²–2B²C²–2C²A² = 4A²B²
A⁴+B⁴+C⁴=4A²B²–2A²B²+2B²C²+2C²A²
A⁴+B⁴+C⁴=2A²B²+2B²C²+2C²A²
A⁴+B⁴+C⁴=2(A²B²+B²C²+C²A²)