What are the three cube roots of 1?(1)^(1/3)?

Answer 1

Just to disambiguate what you're talking about, I think what you want are the solutions to this equation:

x^3 = 1

The reason why I disambiguate is because the cube root is a specific function on your calculator that only has one solution. The equation above has three.

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

x = 1, and
x = [-1 +/- sqrt(1 - 4)]/2
x = [-1 +/- sqrt(-3)]/2
x = (-1/2) +/- (1/2)sqrt(-3)

Since i = sqrt(-1), and sqrt(-3) = sqrt(-1 * -3), then

x = (-1/2) +/- (1/2)sqrt(3)i

So we have one real solution and two imaginary (complex) solutions.

x = {1, -1/2 + (1/2) sqrt(3) i , -1/2 - (1/2) sqrt(3) i }

Answer 2

1

Answer 3

OK ANSWER IS 1, I'LL ANSWER WITH PROOF! suppose x is a variable which u call cube root of 1.So,
(x)3 = 1
which implies, (x)3 - 1 = 0
which implies, [ (x - 1 )] * [ (x)2 - x - 1 ] = 0

Now this means, either ( x - 1 ) = 0 or [ (x)2 - x - 1 ] = 0

Now, if we consider x - 1 = 0 , it implies that x = 1.

And then, if we consider [ (x)2 - x - 1 ] =0,
from the bramhagupta theorem, we po sess two solutions ,
1st one : x = [ 1 + sqrt(-3) ] / 2
2nd one: x = [ 1 - sqrt(-3)] / 2

so, we have three answers, 1 ,[ 1 + sqrt(-3) ] / 2 and
[ 1 - sqrt(-3)] / 2. But the last two numbers are not real numbers and are called complex numbers. So, considering only the real numbers, the solution is 1.

Answer 4

You're not going to find the 3 roots of this by algebra. You find it by using complex numbers. The number -125 can be written as this: 5^3 * exp(i*pi) where i is the square root of 1. You find the three cube roots by going around the complex plane to find the 3 angles that when multiplied by three will give you pi, given that the expression is periodic every 2*pi. So the three roots are going to be: 5*exp(i*pi/3) 5*exp(i*pi) 5*exp(i*5*pi/3) the one in the middle is equal to -5. The other two are most compactly written in how I wrote it. You can expand it out into real and complex parts, but there's no real reason to do so unless you need to work with the two components.

Answer 5

In Complex numbers (where i represents the square root of -1), 1 can be written e^(2*k*pi*i) where e is the base of natural logs and k any integer.

Taking a cube root of a number is the same as dividing its logarithm by 3 (or dividing e'e exponent by 3).

The cube roots of 1 are e^((2*k*pi*i)/3).

Three distinct roots are

e^((2*pi*i)/3), e^((4*pi*i)/3), and e^((6*pi*i)/3)
using k=1, 2, 3.

The third one has the same form as 1, therefore 1 is one of the cube roots of 1 (as is expected).

Since you are styding complex numbers, you can transform the "e" notation back to the standard notation a + bi.

Answer 6

these are 1,w,w^2
their prop are 1+w+w^2=0
w=(1+sqrt(3))/2
w^2=(1-sqrt(3))/2
these roots can be found buy complex number or quadratic equation

Answer 7

i dont think thats even real? well if it is could you tell me i like to learn new things in a while....lol..well hope to here soon!