If n is a positive integer, then $(\sqrt{3}+1{)}^{2\mathrm{n}}-(\sqrt{3}-1{)}^{2\mathrm{n}}$ is 

$(\sqrt{3}+1{)}^{2\mathrm{n}}{=}^{2\mathrm{n}}{\mathrm{C}}_0(\sqrt{3}{)}^{2\mathrm{n}}{+}^{2\mathrm{n}}{\mathrm{C}}_1\left(\sqrt{3}{)}^{2\mathrm{n}-1}{+}^{2\mathrm{n}}{\mathrm{C}}_2\right(\sqrt{3}{)}^{2\mathrm{n}-2}+\dots {+}^{2\mathrm{n}}{\mathrm{C}}_{2\mathrm{n}}(\sqrt{3}{)}^{2\mathrm{n}-2\mathrm{n}}$

and $(\sqrt{3}-1{)}^{2\mathrm{n}}{=}^{2\mathrm{n}}{\mathrm{C}}_0(\sqrt{3}{)}^{2\mathrm{n}}(-1{)}^0{+}^{2\mathrm{n}}{\mathrm{C}}_1(\sqrt{3}{)}^{2\mathrm{n}-1}(-1{)}^1$${+}^{2\mathrm{n}}{\mathrm{C}}_2\left(\sqrt{3}{)}^{2\mathrm{n}-2}\right(-1{)}^2+\dots {+}^{2\mathrm{n}}{\mathrm{C}}_{2\mathrm{n}}\left(\sqrt{3}{)}^{2\mathrm{n}-2\mathrm{n}}\right(-1{)}^{2\mathrm{n}}$

$\therefore (\sqrt{3}+1{)}^{2\mathrm{n}}-(\sqrt{3}-1{)}^{2\mathrm{n}}=2\left[{ }^{2\mathrm{n}}{\mathrm{C}}_1(\sqrt{3}{)}^{2\mathrm{n}-1}\right.{+}^{2\mathrm{n}}{\mathrm{C}}_3\left(\sqrt{3}{)}^{2\mathrm{n}-3}{+}^{2\mathrm{n}}{\mathrm{C}}_5\right(\sqrt{3}{)}^{2\mathrm{n}-5}$$\left.+\dots {+}^{2\mathrm{n}}{\mathrm{C}}_{2\mathrm{n}-1}(\sqrt{3}{)}^{2\mathrm{n}-(2\mathrm{n}-1)}\right]$

which is most certainly an irrational number because of odd powers of $\sqrt{3}$ in each of the terms.