how do you solve x(cube)-1=0? and what are the answers to this equation?

Answer 1

x^3 - 1=0
Since a^3-b^3 = (a-b)(a^2 + b^2 + ab);
x^3 -1 = (x-1)(1+x+x^2)
soln's of (1+x+ x^2):
x={-1+ (1-4)^1/2}/2 ; {-1- (1-4)^1/2}/2
also, sq rt of -1 is defined as IOTA, represented by i.
so, x = (-1+i(sqrt3)}/2 , (-1-i(sqrt3)}/2
the fir term is represented by omega, w
next term is omega sq, w^2
so, cube roots of 1 are: 1, w ,^2

Answer 2

x^3 - 1 = 0

This is a difference of two cubes, which can be factored as :

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

Thus, either (x - 1) = 0, which implies x = 1

or, (x^2 + x + 1) = 0, which by the quadratic formula,
implies x = (-1/2) + i*sqrt(3)/2 or (-1/2) - i*sqrt(3)/2.

Answer 3

Solve x^3 - 1 = 0

First: isolate x^3 on one side by itself > add "1" to both sides....

x^3 - 1 + 1 = 0 + 1

x^3 = 1

Sec: find the cube root of both sides...

V`(x^3) = V`(1)

V`(x*x*x) = V`(1)

x = 1

Answer 4

1 works.

Answer 5

x^3-1= 0 x^3 =1==> x=1 so x^3-1=(x-1)(x ^2+x+1)
the second factor has complex roots x= -1/2 +-(sqrt3)*i/2

Answer 6

X^3-1=0
X^3=1
(X^3)^(1/3)=1^(1/3)
X=1

(the first - the second)(the first square - the first *the second + the second square)

where x: is the first and 1: is the second, and your life will become easy.

so it will be:
(x-1)(x^2-X+1)

Answer 7

x(cube)-1=0
x(cube)=1
then x has to be 1 by inspection.