Question 137778
Problems of the form 
{{{z^2-b^2}}} factor in this fashion
{{{z^2-b^2=(z-b)(z+b)}}}
We can make a substitution to get it into this form.
Let 
{{{z=x^4}}}
{{{b=1}}}
Therefore
{{{z^2-b^2=x^8-1=(x^4-1)(x^4+1)}}}
Similarly
{{{x^4-1=(x^2-1)(x^2+1)}}}
and
{{{x^2-1=(x-1)(x+1)}}}
Substituting all of those in, you get
{{{x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)}}}
The roots of {{{x^2+1}}} are (i, -i).
{{{x^8-1=(x-1)(x+1)(x+i)(x-i)(x^4+1)}}}
Finally if we do a substitution of {{{u=x^2}}} we get,
{{{(x^4+1)=(u^2+1)=(u+i)(u-i)=(x^2+i)(x^2-i)=(x^2+i)(x-sqrt(i))(x+sqrt(i))}}}
{{{(x^4+1)=(x-sqrt(-i))(x+sqrt(-i))(x-sqrt(i))(x+sqrt(i))}}}
When we put that all together we get,
{{{x^8-1=(x-1)(x+1)(x+i)(x-i)(x-sqrt(-i))(x+sqrt(-i))(x-sqrt(i))(x+sqrt(i))}}}
You can check on the square root of i and -i.
The four roots are of the form 
{{{x=0 +- (1/sqrt(2))(1 +- i)}}}