need help with complex number

Hi, the question I'm struggle with is (z^2 + 1)^3 = -1

I have to find all the solutions for this question and the answer should be in Cartesian form

I start the question by powering both side by 1/3

thus the equation become

Z^2 + 1 = -1^(1/3)

since i^2 = -1

Z^2 +1 = (i^2)^(1/3)

from here I sub in Z = X + iY

so the equation become

X^2 + 2XYi - Y^2 = (i^2)^(1/3) + 1

I'm pretty much stuck now and unable to continue the question. Can you guy tell me how to continue this question and get the solution in cartesian form?

(I have a feeling that I did something wrong in the step above

Re: need help with complex number

I think what I would do is write the equation as the sum of two cubes:

$\displaystyle (z^2+1)^3+1^3=0$

Factor:

$\displaystyle ((z^2+1)+1)((z^2+1)^2-(z^2+1)+1)=0$

$\displaystyle (z^2+2)(z^4+z^2+1)=0$

$\displaystyle (z^2+2)(z^2+z+1)(z^2-z+1)=0$

Now you can easily get all six roots in rectangular form.

Re: need help with complex number

I have another tries with the question, thank to your reply I got an idea to move 1 to the other side first

I started by +1 to both side of the equation then powered both side by ^{(1/3)}

the equation become Z^{2} +2 = 0

I then minus -2 to both side

then sub z=x+iY

(x+iY^{2}) = -2

X^{2 }+ 2XYi -Y^{2}= -2

(X^{2} + Y^{2}) - 2XYi = -2

I now have the answer in Cartesian form, however how do I find all the solutions?

thank