A sum of radicals

Compute. $\beta=\alpha+\alpha^2$. $\beta^3=\alpha^3+3\alpha^2\cdot\alpha^2+3\alpha\cdot\alpha^4+\alpha^6=2+3\alpha^4+3\alpha^5+4=6+3\alpha^4+3\alpha^5$.

Reduce mod $\alpha^3=2$: $\alpha^4=2\alpha$, $\alpha^5=2\alpha^2$. So $\beta^3=6+6\alpha+6\alpha^2=6(1+\alpha+\alpha^2)=6(1+\beta)$.

Polynomial. $\beta^3-6\beta-6=0$. Irreducible? Rational root test: $\pm1,\pm2,\pm3,\pm6$: none satisfy. So irreducible of degree $3$.

Conclusion. $\beta$ is algebraic of degree $3$, irrational.

Check. $\beta=\alpha+\alpha^2=\sqrt[3]{2}+\sqrt[3]{4}\approx1.26+1.59=2.85$. $\beta^3-6\beta-6\approx23.1-17.1-6=0$. ✓