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I have a question about Cube root Concept:?

(8)^(1/3) = 2

(- 8)^(1/3) = - 2 <----- is this true??????

if so, why wolframalpha doesn't show the real root? is there a trick somewhere?

http://www.wolframalpha.com/input/?i=%28%E2%80%A6

If I have an equation as 3x^3 - 2 = 0

the roots are as the following:

3 * ( x^3 - 2/3) ) = 0

x^3 - (2/3) = 0

(x - (2/3)^(1/3)) * (x^2 + (2/3)^(1/3)x + (2/3)^(2/3)) = 0

(x - (2/3)^(1/3)) = 0 ----> x = (2/3)^(1/3)

(x^2 + (2/3)^(1/3)x + (2/3)^(2/3)) = 0 ---> quadratic equation and I came up with:

- (2/3)^(1/3) * [ (1+/-i√(3) )/ 2 ]

BUT, the wolfram alpha shows in the following link:

http://www.wolframalpha.com/input/?i=3x%%E2%80%A6

how they have gotten the complex roots?

if I use my TI-89 Calculator, it won't show the imaginary root for the equation 3x^3 - 2 = 0 ? any idea?

thank you :)

Update:

the input above website is not true. here is the other one:

http://www.wolframalpha.com/input/?i=%28-8%29%5E%2...

Update 3:

thank you so much...but actually, if we press to approximate the answer, i will match them.....but i am just wondering of the form that they have. can you tell?

1 Answer

Relevance
  • Anonymous
    9 years ago
    Favorite Answer

    The highest power of (x) in this case is 3; therefore, there must be 3 roots to the equation. One root is the obvious (2/3)^(1/3). As the other two roots are not apparent from the equation we understand that the remaining two roots must be a complex conjugate pair. For some reason the links in your description are not opening properly. However, I have just confirmed your use of the quadratic formula on the function of (x) and have retrieved the same complex roots. Therefore, as the mathematics is accurate your solutions must be right. WolframAlpha may have made a mistake.

    The problem with your calculator is more simple, visit the following link it is a very helpful and succinct site for TI-89 users: http://mathbits.com/mathbits/tisection/voyage/home...

    I just saw the new links and noticed that the so-called complex roots do not even include the complex integer (i) and so I would not believe that answer. The real root of square root of (-8) is (-2) again WolframAlpha has slipped up. Right about now I would question the validity of the data on this site. Maybe try some different sites out to confirm your answer but again your methodology is accurate so I wouldn't worry.

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