how to find all real solutions of an equation

Question

x^6-2x^3-3=0

Answer 1

It looks easy to factor

(x^3-3)(x^3+1)

the first three roots are the cube root of 3

the second three roots are the cube root of -1, which are -1 itself!

They’re all six real roots.

Answer 2

x⁶ - 2x³ - 3 = 0

(x³ - 3)(x³ + 1) = 0

x³ - 3 = 0    or    x³ + 1 = 0

x³ = 3

x =  ³√3

x = 1.4422

 

x³ + 1 = 0

x³ = -1

x = ³√-1

x = i, not a real solution