Prove the sum of $sqrt{3} + sqrt[3]{4}$ is Irrational number

The following should allow you to generalize to combinations of a quadratic and a cubic irrational in general:

Every number of the form $alpha=a+bsqrt 3$ with $a,binBbb Q$ is a root of a quadratic polynomial because $(a+bsqrt3)^2=a^2+3b^2+2absqrt 3$ and so $alpha$ is a solution of of $$X^2-2aX+a^2-3b^2=0. $$ If $sqrt 3-sqrt[3]4$ is rational, it follows that the above applies to $alpha:=sqrt[3]4$. But we also know that it is a solution of the $X^3-4=0$. Hence it is also a solution of $$ 0=(X^3-4)-(X+2a)cdot(X^2-2aX+a^2-3b^2)=(ldots)cdot X+(ldots)$$ and therefore rational - contradiction. (Verify that a least one of the "$(ldots)$" is non-zero!)

Let $x=sqrt{3}+sqrt[3]{4}$ so $0=(x-sqrt{3})^3-4=x^3-3x^2sqrt{3}+9x-3sqrt{3}-4$. If $xinBbb Q$, $sqrt{3}=frac{x^3+9x-4}{3(x^2+1)}inBbb Q$, a contradiction.