How do you solve this cubic equation?

Question

x^3-1 = 0

Answer 1

(x-1)(x^2+x+1)=0
x-1=0 x=1
for x^2+x+1 using the Quadratic formula
x=[-1+/-rt(-3)]/2

so the three roots are
x=1,(1+irt3)/2,(1-irt3)/2

Answer 2

f(x) = x^3-1
we know by inspection
f(1) = 0
so (x-1) is a factor

by synthetic division
f(x) = (x-1)(x^2+x+1) = 0

x =1 or

x^2+x + 1 = 0 ..1

this can be solved for 2 additional values of x

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

the

Answer 3

One obvious solution is x=1
Divide x^3-1 by x-1 which is x^2+x+1
Solutions of this quadratic eq are
x= -1/2 +or - *sqrt(3)*i/2
where i stand for the complex root of -1

Answer 4

(x-1)(x^2+x+1)=0
as x-1=0 x=1
x^2+x+1 & using the eq
x=[-1+/-rt(-3)]/2

so the three roots are
x=1,(1+irt3)/2,(1-irt3)/2

Answer 5

Write it as x^3=1, the roots could then be -1,-1and 1 or simply just 1.

Answer 6

long division yeilds
x^3 -1 / (x-1) = x^2 + 1 (x+i) (x-i) (x-1)

Answer 7

x^3=1
x=1

Answer 8

X^3=1
cube root of x =1
There fore x = 1
There you have the solution