Prove $x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ is irrational

Let $y = \sqrt[100]{\sqrt{3} + \sqrt{2}}$. Then $x = y + {1 \over y}$. Suppose $x$ were some rational number $q$. Then $y^2 - qy + 1 = 0$. This means ${\mathbb Q}(y)$ is a field extension of ${\mathbb Q}$ of degree two, and every rational function of $y$ is in this field extension. This includes $y^{100} = \sqrt{3} + \sqrt{2}$, and $y^{-100} = \sqrt{3} - \sqrt{2}$. So their sum and difference is also in ${\mathbb Q}(y)$. Hence ${\mathbb Q}(\sqrt{2},\sqrt{3}) \subset {\mathbb Q}(y)$. But ${\mathbb Q}(\sqrt{2},\sqrt{3})$ is an extension of ${\mathbb Q}$ of degree 4, a contradiction.

You can make the above more elementary by showing successive powers of $y$ are always of the form $q_1y + q_2$ with $q_1$ and $q_2$ rational and eventually showing some $q_3\sqrt{2} + q_4\sqrt{3}$ must be rational, a contradiction.

HINT $\rm\quad \alpha+\bar\alpha,\ \alpha\:\bar\alpha\:\in \mathbb Q\ \Rightarrow\ \alpha^{\:n}+\bar\alpha^{\:n}\in \mathbb Q\ $ by induction, since

$$\rm\ \alpha^{\:n+1}\!+\:\bar\alpha^{\:n+1}\ =\ (\alpha+\bar\alpha)\ (\alpha^n\!+\:\bar\alpha^n)-\alpha\:\bar\alpha\ (\alpha^{\:n-1}\!+\:\bar\alpha^{\:n-1})$$

In your case $\rm\ \alpha\:\bar\alpha = 1\in\mathbb Q\ $ so $\rm\ \alpha + \bar\alpha\ \in\mathbb Q\ \Rightarrow\ \alpha^{\:100} + \bar\alpha^{\:100} =\ 2\:\sqrt{3}\in\mathbb Q\ \Rightarrow\Leftarrow$

Look up Lucas-Lehmer sequences to learn more about such power sums of algebraic roots.

If $y + y^{-1} = x$, then $y^n + y^{-n}$ is a polynomial $F_n(x)$ with integer coefficients in $x$. In particular, if $x$ is rational, then so is: $$F_{100}(x) = y^{100} + y^{-100} = (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) =2 \sqrt{3}.$$

The polynomials $F_{n}(x)$ are (up to scaling and renormalization) the Chebyshev polynomials. One has $F_0(x) = 2$, $F_1(x) = x$, $F_2(x) = x^2 - 2$, and $F_{n}(x) = x F_{n-1}(x) - F_{n-2}(x)$.