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12. If a+b+c 11, ab+bc+ca-20, then the value of a b+c3 -3abc is:(a) 121(b) 341(c) 671 |
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Answer» a + b + c = 11 and ab+bc+ca = 20 (a+b+c)(a+b+c) = 121 a*a+b*b+c*c+2ab+2bc+2ca = 121 a*a+b*b+c*c+40 = 121 a*a+b*b+c*c = 81 a*a*a+b*b*b+c*c*c-3abc = (a+b+c)(a*a+b*b+c*c-ab-bc-ca) = 11(81-20) = 11*61 = 671 |
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