Question

Solve?

Answer 1

since...we will be having two values of x...which one...shuld we use

Answer 2

maybe try one and then the other. roots are complex, right?

Answer 3

in fact i think they are the imaginary cube roots of 1

Answer 4

so it probably doesn't matter which one you use

Answer 5

@satellite73 for that.. the equation must be x^2+x+1=0

Answer 6

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

Answer 7

\(e^{\frac{\pi}{3}i}\) and its conjugate, which also in this case happens to be its reciprocal
\(e^{-\frac{\pi}{3}i}\)

Answer 8

@bhaweshwebmaster that is for the cube roots of 1

Answer 9

\[x^3=-1\]
\[x^3+1=0\]
\[(x+1)(x^2-x+1)=0\] etc

Answer 10

@satellite73 i cant understand..wat ur saying right nw....))

Answer 11

x^2-x+1=0

Answer 12

did you understand my very last post? showing that you have the two imaginary cube roots of \(-1\) ?

Answer 13

there will be two values for x (complex

Answer 14

Yup....i knw cube root of unity

Answer 15

in this case not the cube roots of 1, but the cube roots of \(-1\)