Prove that $\sqrt{10} - \sqrt6 - \sqrt5 + \sqrt3$ is irrational

Question

I tried the methods shown in Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares? but I cannot extend them well into 4 numbers.

Answer 1

Let $a=\sqrt6+\sqrt5$ and $b=\sqrt{10}+\sqrt3$. You want to show that $b-a\notin\mathbb Q$.

Notice that $$ \begin{align*} a^2&=11+2\sqrt{30}\\ b^2&=13+2\sqrt{30} \end{align*} $$ hence $$b^2-a^2=(b-a)(b+a)=2$$ and $b+a=\frac2{b-a}$.

Now if $b-a\in\mathbb Q$, then also $b+a=\frac2{b-a}\in\mathbb Q$ and $$b=\frac{(b-a)+(b+a)}2\in\mathbb Q,$$ a contradiction. (To prove that $b$ is irrational you can use methods from the thread you linked.)

Answer 2

$$\alpha=\sqrt{10}-\sqrt{6}-\sqrt{5}+\sqrt{3} = \left(\sqrt{2}-1\right)\left(\sqrt{5}-\sqrt{3}\right)$$ Hence: $$ \alpha^2 = \left(3-2\sqrt{2}\right)\left(8-2\sqrt{15}\right)$$ and since $\sqrt{15}$ does not belong to $\mathbb{Q}(\sqrt{2})$, $\alpha^2$ is not a rational number, hence $\alpha\not\in\mathbb{Q}$.