InterviewSolution
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InterviewSolution
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\(^3\sqrt{a}\) + \(^3\sqrt{b}\) + \(^3\sqrt{c}\) = 0 then (a + b + c)3 =................(A) 3 abc (B) 9 abc (C) 27 abc (D) 81 abc |
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Answer» Correct option is (C) 27 abc We have \(\sqrt[3]a+\sqrt[3]b+ \sqrt[3]c=0\) \(\Rightarrow\) \(a^\frac13+b^\frac13+c^\frac13=0\) \(\Rightarrow\) \(a^\frac13+b^\frac13=-c^\frac13\) ___________(1) \(\Rightarrow\) \((a^\frac13+b^\frac13)^3=(-c^\frac13)^3\) (By cubing both sides) \(\Rightarrow(a^\frac13)^3+(b^\frac13)^3+3a^\frac13b^\frac13(a^\frac13+b^\frac13)=-(c^\frac13)^3\) \((\because(a+b)^3=a^3+b^3+3ab(a+b)\,\&\,(ab)^3=a^3b^3)\) \(\Rightarrow a+b+3\,a^\frac13\,b^\frac13\times-c^\frac13=-c\) (From (1)) \(\Rightarrow\) a+b+c \(=3\,a^\frac13\,b^\frac13c^\frac13\) \(\Rightarrow\) \((a+b+c)^3\) \(=(3\,a^\frac13\,b^\frac13c^\frac13)^3\) (By cubing both sides) \(\Rightarrow\) \((a+b+c)^3=27\,abc\) \((\because(ab)^m=a^mb^m)\) Correct option is (C) 27 abc |
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