$\sqrt{2} + \sqrt[3]{3}$ is irrational ?
These are my steps -
$\sqrt{2} + \sqrt[3]{3} = a$
$3 = (a-\sqrt{2})^{3}$
$3 = a^{3} -3a^{2}\sqrt{2} + 6a -2\sqrt{2}$
$3a^{2}\sqrt{2}+2\sqrt{2} = a^{3}+6a-3$
$\sqrt{2}(3a^{2}+2) = a^{3}+6a-3$
Then, $\sqrt{2}$ in the left side is irrational , and mulitply irratinal with rational is irrational. The right side is rational. So, $irrational \neq rational$.
This is a good proof ?
Taking powers of $\alpha=\sqrt2+\sqrt[\large3]{3}$ and putting them into matrix form, we get $$ \begin{bmatrix} \alpha^0\\\alpha^1\\\alpha^2\\\alpha^3\\\alpha^4\\\alpha^5\\\alpha^6 \end{bmatrix} = \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&1&0&0&0\\ 2&0&0&2&1&0\\ 3&2&6&0&0&3\\ 4&12&3&8&12&0\\ 60&4&20&15&3&20\\ 17&120&90&24&60&18 \end{bmatrix} \begin{bmatrix} 1\\2^{1/2}\\3^{1/3}\\2^{1/2}3^{1/3}\\3^{2/3}\\2^{1/2}3^{2/3} \end{bmatrix}\tag{1} $$ We can use the method from this answer to get a vector perpendicular to all the columns in the matrix above: $$ \begin{bmatrix} 1\\-36\\12\\-6\\-6\\0\\1 \end{bmatrix}^{\large T} \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&1&0&0&0\\ 2&0&0&2&1&0\\ 3&2&6&0&0&3\\ 4&12&3&8&12&0\\ 60&4&20&15&3&20\\ 17&120&90&24&60&18 \end{bmatrix}=0\tag{2} $$ $(1)$ and $(2)$ imply that $$ \alpha^6-6\alpha^4-6\alpha^3+12\alpha^2-36\alpha+1=0\tag{3} $$ $(3)$ says that $\alpha$ is an algebraic integer. A rational algebraic integer must be an integer. However, $1\lt\sqrt2\lt\frac32$ and $1\lt\sqrt[\large3]3\lt\frac32$, thus $2\lt\alpha\lt3$. Therefore, $\alpha$ must be irrational.