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

$(\sqrt{3}+1{)}^{2n}-(\sqrt{3}-1{)}^{2n}=[(\sqrt{3}+1{)}^2{]}^n-[(\sqrt{3}-1{)}^2{]}^n$$=(3+1+2\sqrt{3}{)}^n-(3+1-2\sqrt{3}{)}^n$$=(4+2\sqrt{3}{)}^n-(4-2\sqrt{3}{)}^n$$=2^n[(2+\sqrt{3}{)}^n-(2-\sqrt{3}{)}^n]$ Let $(2+\sqrt{3}{)}^n=I+f$ and $(2-\sqrt{3}{)}^n=f^{\prime }$.Using Binomial Expansion:$(2+\sqrt{3}{)}^n=(\genfrac{}{}{0ex}{}{n}{0})2^n+(\genfrac{}{}{0ex}{}{n}{1})2^{n-1}(\sqrt{3})+(\genfrac{}{}{0ex}{}{n}{2})2^{n-2}(\sqrt{3}{)}^2+\dots$$(2-\sqrt{3}{)}^n=(\genfrac{}{}{0ex}{}{n}{0})2^n-(\genfrac{}{}{0ex}{}{n}{1})2^{n-1}(\sqrt{3})+(\genfrac{}{}{0ex}{}{n}{2})2^{n-2}(\sqrt{3}{)}^2-\dots$ Subtracting the two:$(2+\sqrt{3}{)}^n-(2-\sqrt{3}{)}^n=2\left[\left(\genfrac{}{}{0ex}{}{n}{1}\right)2^{n-1}(\sqrt{3})+\left(\genfrac{}{}{0ex}{}{n}{3}\right)2^{n-3}(\sqrt{3}{)}^3+\dots \right]$$=2\sqrt{3}\left[\left(\genfrac{}{}{0ex}{}{n}{1}\right)2^{n-1}+\left(\genfrac{}{}{0ex}{}{n}{3}\right)2^{n-3}(3)+\dots \right]$ This is of the form $2\sqrt{3}\times (\text{Integer})$.Since $n$ is a positive integer, the expression results in a multiple of $\sqrt{3}$, which is irrational.