## College Algebra 7th Edition

$3520\sqrt{2}y^{3}$
We find the term with $y^3$ in the expansion of $(\sqrt{2}+y)^{12}$ by using Binomial Theorem coefficients: $\left(\begin{array}{l} 12\\9\end{array}\right)(\sqrt{2})^{9}y^{3}=220*(\sqrt{2})^8\sqrt{2}y^3=220*16\sqrt{2}y^3=3520\sqrt{2}y^{3}$