Show me how to solve x^4-1=0?
it is a difference of two squares.
(x^2-1)(x^2+1)=0
which means x^2-1=0 or x^2+1=0
so in the first case,
x^2=1
x=+-1
and in the second case
x^2=-1
x= i where i is equal to sqrt(-1)
(x^2-1)(x^2+1)=0
which means x^2-1=0 or x^2+1=0
so in the first case,
x^2=1
x=+-1
and in the second case
x^2=-1
x= i where i is equal to sqrt(-1)