I know about the square root of -1, what can you tell me about the cube root of-1?

Answer 1

The cube root of -1 is an Algebraic number, which means there is a polynomial f(x) with rational cooefficents of finite degree to which (-1)^(1/3) is a root.

x^3 + 1 = 0 is such a polynomial.

There are no real numbers that are the cube root of -1, however there ARE complex numbers that are the cube root of -1.

Three in fact, yes, all three numbers are the cube root of -1.

1/2 + [sqrt(3) / 2]*i is one,
1/2 - [sqrt(3) / 2]*i is the second,
-1 itself is the last one.

Proof: -1 is trivial.

[1/2 + [sqrt(3) / 2]*i ]^3
= [1/2 + [sqrt(3) / 2]*i ] * [1/2 + [sqrt(3) / 2]*i ] * [1/2 + [sqrt(3) / 2]*i ]
= [-1/2 + [sqrt(3) / 2]*i ] * [1/2 + [sqrt(3) / 2]*i ]
= -1

Try the last one out yourself using FOIL, and rememebr i^2 = -1

Hope this helps.
- Jack

(EDIT: Replaced sqrt(2) with sqrt(3), now it's correct)

Answer 2

Some of you are confusing powers and exponents with roots. The ath root of b = b to the power of the reciprical of a.
i.e., the cube root is (-1)^(1/3) and NOT (-1)^(3). It is not -1.

You could consider the cube root of -1 to be the square root of -1 to the power 2/3.
Also, .5 + .5*(3)^.5*(-1)^.5, according to my TI-89.

I don't believe that it has any special symbol or mathematical signifigance as does sqrt(-1), if that's what you're wanting to know.

Answer 3

The cube root of -1 is -1. -1 cubed is -1.

Answer 4

cube root of-1 is -1, because (-1)^3 = -1

Answer 5

-1

Answer 6

-1

Answer 7

There are three cube roots: -1 (of course), 1/2 + i x sqrt(3)/2, 1/2 - i x sqrt(3)/2.

Answer 8

let us consider that the cube root of -1 is a complex number z
therfore -1=z^3
we can treplace z with x+iy
or -1=(x+iy)^3
or -1= x^3+(iy)^3+3yx^2i+3x^2yi
or -1=x^3 - y^3 + (3x^2y + 3xy^2)i
since there is no imaginary part
(3x^2y + 3xy^2)i=0
and
-1=x^3 - y^3
or (x-y) (x^2 + y^2 + xy)= -1
therfore either x-y= -1
or (x^2 + y^2 + xy)=0\
on solving we see the roots of
we get -1/2 + i root3/2 and -1/2- iroot3/2
hence we have three solutions
-1 , w [ -1/2 + i root3/2] (omega) and w^2 [-1/2- iroot3/2](omega square)

Answer 9

square root of -1 is denoted as "i" and forms the basis of complex number system

cube root of -1 is -1, as well as two complex numbers:
(1-sqrt(3)*i)/2 and (1+sqrt(3)*i)/2

Answer 10

It is -1. There is no problem with odd numbered roots of negative numbers. The problem is that imaginary numbers are defined as being too complex for elementary mathematics, as well as complex numbers. Most math, prior to calculus limits itself to the use of real numbers, such that any function which yields an imaginary number in its answer is considered an illegal operation under the Real number set.