Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 636: 12

$99+70\sqrt{2}$

We use the values from the 6th row of Pascal's Triangle to find the coefficients: $(1+\sqrt{2})^{6}=1^{6}+6*1^{5}*\sqrt{2}+15*1^{4}*2+20*1^{3}*2\sqrt{2}+15* 1^{2}* 4+6* 1^1* 4\sqrt{2}+2^{3}1^0=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8=99+70\sqrt{2}$