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If a+b+c=6 and ab+bc+ca=11 find the value of a^3+b^3+c^3-3abc |
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Answer» Step-by-step explanation: Given a+B+c = 6 , ab + bc + CA = 11 ( a + b + c )² = a² + b² + c² + 2 (ab + bc + ca) ⇒ (6)² = a² + b² + c² + 2(11) ⇒ a² + b² + c² = 36-22 = 14 now a³ + b³ + c³ -3abc= ( a + b + c) (a² + b² + c² - ab - bc - ca ) ⇒ (6)(14 - 11) = 6. 3 = 18 |
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