Let E =Q(α,ω), α = 2^{1/3}, ω =e^{2}^{π}^{i}^{/3}, (Qis the field of rationals)

Is E a simple extension of Q?

My attempt:

Let β = α + ω, find a monic irreducible polynomial, p inQ[x] with root β.

p should have degree 6, implying E =Q(β)

β^{3}= (α + ω)^{3}

= α^{3}+ 3α^{2}ω + 3αω^{2}+ ω^{3}

= 2 + 3α^{2}ω + 3αω^{2}+ 1

= 3αω(α + ω) + 3

= 3αωβ + 3

β^{3} 3 = 3αωβ

(β^{3} 3)^{3}= (3αωβ)^{3}

(β^{3})^{3}+ 3(β^{3})^{2}(-3) + 3(β^{3}) (-3)^{2}+ (-3)^{3}= 3^{3}α^{3}ω^{3}β^{3}

β^{9}- 9β^{6}+ 27β^{3} 27 = 9(2)(1)β^{3}

= 18 β^{3}

β^{9}- 9β^{6}+ 9β^{3} 27 = 0

However, if I define p(x) = x^{9} 9x^{6}+ 9x^{3} 27 = 0, my calculator tells me p(2^{1/3}+e^{2}^{π}^{i}^{/3}) ≠ 0.

Also I cant see how to reduce p such that p = fg with f,g inQ[x], deg(f) = 6, and f(β) = 0.