I have a question about Cube root Concept:?

Question

(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=%28-8%29%5E%281%2F3%29

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%5E3+-+2+%3D+0+

how they have gotten the complex roots?

Colin · Accepted Answer

"(- 8)^(1/3) = - 2 <----- is this true?????? "

As -2*-2*-2 = - 8, I think so.

Let&#39;squestion · Answer

General cube roots of -1 are given by are given by 
cos [(2p*pi+pi)/3] + i*sin [(2p*pi+pi)/3], p = 0 and p =2 give two complex conjugate roots and p = 1 gives cos pi + sin pi = -1 as the real root.

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I have a question about Cube root Concept:?

2 Answers