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Question
If a1/3 + b1/3 + c1/3 = 0, then
Options
a + b + c = 0
(a + b + c)3 =27abc
a + b + c = 3abc
a3 + b3 + c3 = 0
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Solution
Given `a^(1/3) +b^(1/3) +c^(1/3) = 0`
Using identity `a^3 +b^3 +c^3 = 3abc` we get
Here `a= a^(1/3) ,b=b^(1/3) , c = c^(1/3) `
`(a^(1/3))^3 + (b^(1/3))^3 +(c^(1/3))^3 = 3 xx a^(1/3) xx b^(1/3) xx c^(1/3)`
`(3sqrta)^3 +(3sqrtb)^3 +(3sqrtc)^3 =3 xx 3sqrta xx 3sqrtb xx3sqrt c`
`a+b+c = 3 xx 3sqrt a xx 3sqrtb xx 3sqrtc`
Taking Cube on both sides we get,
`(a+b+c)^3 = (3xx 3sqrta xx 3sqrtb xx 3sqrtc)^3`
`(a+b+c)^3 = 27abc`
Hence the value of `a^(1/3) +b^(1/3) +c^(1/3) = 0` is `(a+b+c)^3 = 27abc` .
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