If a + b + c = 6, a² + b² + c² = 14 and ab + bc+ ca = 11, then what is the value of a³ + b³ +c³ –3abc?
Given – a + b + c = 6 a² + b² + c² = 14 ab + bc+ ca = 11 then a³ + b³ +c³ –3abc = (a+ b+ c) (a² + b² + c² -ab-bc-ca) = 6 (14-11) =6 ×3=18
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