Mathematics - Binomial Theorem Question with Solution | TestHub

Let $x=(\sqrt{2}+1{)}^6$. Consider $y=(\sqrt{2}-1{)}^6$. Note that $0<\sqrt{2}-1<1$, so $0<y<1$.

Using the binomial theorem, $x+y=(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6=2[(\genfrac{}{}{0ex}{}{6}{0})(\sqrt{2}{)}^6+\left(\genfrac{}{}{0ex}{}{6}{2}\right)(\sqrt{2}{)}^4+(\genfrac{}{}{0ex}{}{6}{4})(\sqrt{2}{)}^2+\left(\genfrac{}{}{0ex}{}{6}{6}\right)]=2[8+15(4)+15(2)+1]=2[8+60+30+1]=2[99]=198$.

Since $x+y=198$ and $0<y<1$, the greatest integer less than or equal to $x$ is 197.

Final Answer: The final answer is $\boxed{197}$