Prove that $\sqrt{2} + \sqrt[3]{3}$ is irrational $\sqrt{2} + \sqrt[3]{3}$ 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 ?
 A: 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.
