Assume q(x)=x³+a₂x²+a₁x+a₀.
Let y be a zero of q(x). Using synthetic division:
y | 1 a₂ a₁ a₀
1 y a₂y+y² a₁y+a₂y²+y³
1 a₂+y a₁+a₂y+y² | a₀+a₁y+a₂y²+y³=0 remainder
Now replace y with c+d(∛2):
a₀+a₁(c+d(∛2))+a₂(c+d(∛2))²+(c+d(∛2))³=0,
a₀+a₁(c+d(∛2))+a₂(c²+2cd(∛2)+d²(2^⅔))+(c³+3c²d(∛2)+3cd²(2^⅔)+2d³)=0,
a₀+a₁c+a₂c²+c³+2d³+d(∛2)(a₁+2a₂c+3c²)+d²(2^⅔)(a₂+3c)=0.
To remove the irrationals:
a₁+2a₂c+3c²=0, a₂+3c=0, a₂=-3c, a₁-6c²+3c²=0, a₁=3c², if we ignore the trivial case of d=0.
To find a₀, substitute for a₁ and a₂:
a₀+a₁c+a₂c²+c³+2d³=0, a₀+3c³-3c³+c³+2d³=0, a₀=-c³-2d³.
q(x)=x³-3cx²+3c²x-c³-2d³,
which is (x-c)³-2d³=(x-c-d(∛2))((x-c)²+(c+d(∛2))(x-c)+(c+d(∛2))²).