If a²+b²+c²=0 then prove a⁶+b⁶+c⁶=3a²b²c².
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c=-a-b → c³=-a³-b³–3ab(a+b) → a³+b³+c³+3ab(a+b)=0 → 3ab(a+b)=-6 → ab(a+b)=-2 (I) a²+b²+c²=a²+b²+(a+b)²=6 → 2a²+2ab+2b²=6 → a²+ab+b²=3 (II).
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a⁶+b⁶+c⁶=3a²b²c².
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