Integer related to expansion

There is another approach: Let $(\sqrt{2}+1)^6 = I+F$ ,where $I$ is integral part and $F$ is fractional part.

Let $f$ = $(\sqrt{2}-1)^6$ . We have, $\sqrt{2}-1 = 1/(\sqrt{2}+1)$ implies that $0<f<1$

Also $I+F+f = (\sqrt{2}+1)^6 +(\sqrt{2}-1)^6 = 198$

Hence, $F+f=198-I$ is an integer. But $0<F+f<2$

Therefore, $F+f=1$ , and thus, $\boxed{ I=197}$

Expanding using the binomial expansion formula , we have that

$(2^{\frac{1}{2}} + 1)^{6} = (2^{\frac{1}{2}})^{6} + 6(2^{\frac{1}{2}})^{5} + 15(2^{\frac{1}{2}})^{4} + 20(2^{\frac{1}{2}})^{3} + 15(2^{\frac{1}{2}})^{2} + 6(2^{\frac{1}{2}}) + 1 =$

$8 + 24\sqrt{2} + 60 + 40\sqrt{2} + 30 + 6\sqrt{2} + 1 = 99 + 70\sqrt{2}$ .

Now $(70\sqrt{2})^{2} = 4900 * 2 = 9800 \lt 9801 = 99^{2}$ , and so $70\sqrt{2} \lt 99$ .

Also $70\sqrt{2} \gt 70\sqrt{1.96} = 70*1.4 = 98$ , so

$99 + 98 \lt 99 + 70\sqrt{2} \lt 99 + 99 \Longrightarrow 197 \lt 99 + 70\sqrt{2} \lt 198$ ,

and thus $\lfloor (\sqrt{2} + 1)^{6} \rfloor = \boxed{197}$ .