Find the real and complex solutions to the equation x(3rd power) +1=0?

Answer 1

I am writing x3 for x^3 and 3x for 3 into x

x3 + 1 = 0
or, (x+1)(x2-x+1)=0
reat root is x = -1
solving x2 -x + 1 = 0
x = [1+i sqrt(3)]/2 and x = [1- i sqrt(3)]/2
these two are imaginary roots..

Answer 2

x^3 +1= 0 so x= cubic root of-1 =-1
factoring x^3 +1 ( (x+1)( x^2 -x +1)
The roots of the 2nd factor are 1/2 +(sqrt3)/2 *i and 1/2 -sqrt(3)/2*i
The real solution is -1

Answer 3

x^3=-1+0.i

has 3 roots
arqument= 180+k 360
roots=1(cos (60+k120)+isin(60+k120)
for k=0 the first root is 1/2+i sqrt(3)/2
for k=1 the second root 1(-1+i0)
for k=2 the third root 1(cos300+isin300)=1/2-isqrt(3)/2

Hint: use De Moivre rule

Answer 4

x^3 +1 = 0
x^3 = -1
x = cube root of -1
Therefore x = -1

This function / graph has 3 real and equal roots (solutions), ie -1.