Question

so what is the cubic root of -1
sound like an easy answer that it's -1
but I researched a little bit and found it on http://www.wolframalpha.com/input/?i=cubic+root+-1
that it is not -1
can any body explain

Answer 1

You might know that every nth degree polynomial has exactly n roots. That's true of the polynomial x^3+1, whose roots correspond to the solutions of the equation x^3=-1. This implies that there are three cube roots of -1, just like there are two square roots of any number (positive and negative). One of the cube roots of -1 is -1, like you said, and the one wolfram alpha gave is another one. Here's two explanations, the first one for the specific case you gave and involving only algebra, and the second one for the general case but involving some fancier math.

1. You can factor the polynomial as a sum of cubes and get the equation (x+1)(x^2-x+1) = 0. x=-1 is clearly a solution, but so are both of the solutions to x^2-x+1 = 0. Using the quadratic formula yields 0.5+0.866i and 0.5-0.866i. There are the three cube roots of -1.

2. By De Moivre's Theorem, a complex number cos(theta)+i*sin(theta) raised to the nth power is cos(n*theta)+i*sin(n*theta). -1 in complex polar form is cos(pi)+i*sin(pi). The period of sine and cosine is 2*pi, so cos(pi)+i*sin(pi) is equivalent to cos(3*pi)+i*sin(3*pi) and cos(-pi)+i*sin(-pi). Applying De Moivre's Theorem backwards on each of these yields cos(pi/3)+i*sin(pi/3), cos(pi)+i*sin(pi), and cos(-pi/3)+i*sin(-pi/3). You can verify that these three numbers are equivalent to the numbers found in part 1.