Question

x^6-3x^2-x+1=0 prove that it has two imaginary roots


pls help me by the problem please please

Answer 1

que
two solution were found
1:-x=1
2:-x=2
step by step solotion
step 1:-

Equation at the end of step:-
(((x6)-(3.(x4)))+3x²)-1=0
step:-2
(((x6)-3x4)+3x²)-1=0
checking for a perfect cube
x6-3x4+3x²-1 is not a perfect cube