Edward C. answered 04/21/15
Tutor
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Caltech Grad for math tutoring: Algebra through Calculus
This is a "quadratic in disguise"
Let y = x^3. Then y^2 = x^6 and the equation is
y^2 - 2y - 3 = 0
(y - 3)(y + 1) = 0
y = 3 or y = -1
Remember y = x^3 so
x^3 = 3 or x^3 = -1
x = 3√3 or x = -1
Check: (3√3)^6 - 2(3√3)^3 - 3 = 9 - 6 - 3 = 0
(-1)^6 - 2(-1)^3 - 3 = 1 + 2 - 3 = 0