Show that $4^{1/3}+10^{1/2}$ is irrational.
I start by assuming it to be rational and want to come to a contradiction $$ 4^{1/3}+10^{1/2} = r \\ \Rightarrow 4^{1/3} = r - 10^{1/2} \\ \Rightarrow 4 = (r-10^{1/2})^{3} \\ \Rightarrow 4 = r^{1/3} - 10^{1/3} + 3r^2*10^{1/2} - 30r $$ Now I want to separate $10^{1/2}$ to one side and show the terms with $r$ is rational and thus a contradiction but what do I do with the $10^{1/3}$ on RHS?
On
Just to make @lulu's point explicit: since $0=4^{2/3}-2r4^{1/3}+r^2-10$, if $r$ is rational $4^{1/3}$ has algebraic degree $\le2$, but it doesn't.
On
It is known that for any $a,b\in\mathbb Q^*$, the number $a\sqrt[3]4+b\sqrt{10}$ is a primitive element of the extension $\mathbb Q(\sqrt[3]4,\sqrt{10})$ which is clearly of degree $6$ so $a\sqrt[3]4+b\sqrt{10}$ is of degree $6$ for all pair $a,b$ of non-zero rational. Making $a=b=1$ we are done.
Since $\sqrt[3]{4}$ is a root of the polynomial $f(x) = x^3 - 4$, $r \stackrel{def}{=}\sqrt[3]{4} + \sqrt{10}$ is a root of $$g(x) = f(x-\sqrt{10})f(x+\sqrt{10}) = x^6-30x^4-8x^3+300x^2-240x-984$$ By rational root theorem, if $r = \frac{p}{q}$ is a rational root of $g(x)$ for coprime integers $p$ and $q$, then $q$ is a factor of $1$. This forces $q = \pm 1$ and $r$ to be an integer. Numerically, $$4 < r = \sqrt[3]{4} + \sqrt{10} \sim 4.749678712136578 < 5$$ and $r$ is not an integer. This means $r$ cannot be a rational root of $g(x)$ and hence $r$ is irrational.
Update
For an alternate proof which doesn't involve the horrible sextic polynomial, just expand the equality $f(r - \sqrt{10}) = 0$. You will get
$$\begin{align} & r^3-3\sqrt{10}r^2+30r-10\sqrt{10} - 4 = 0\\ \iff & (r^3 + 30r - 4) - (3r^2+10)\sqrt{10} = 0\\ \implies & \sqrt{10} = \frac{r^3 + 30r - 4}{3r^2 + 10}\end{align}$$ If $r$ is rational, last equality tell us $\sqrt{10}$ will be rational too. This contradicts with the known fact that $\sqrt{10}$ is irrational. As a result, $r$ cannot be rational.