Answer
$(x^{2}+y)^{4} $
Work Step by Step
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
-------------------
There are 5 terms in the expression, so we aim for n=4.
The first and last (1st and 5th) terms,
$x^{8}=(x^{2})^{4}$ and $y^{4}=(y)^{4}$
lead us to check whether the expression equals $(x^{2}+y)^{4} $
$\left(\begin{array}{l}
4\\
1
\end{array}\right)=\left(\begin{array}{l}
4\\
3
\end{array}\right)=4,\ \displaystyle \left(\begin{array}{l}
4\\
2
\end{array}\right)=\frac{4(3)}{1(2)}=6$
2nd term: $\left(\begin{array}{l}
4\\
1
\end{array}\right)(x^{2})^{3}y^{1}=4x^{6}y\qquad $..OK
3rd term: $\left(\begin{array}{l}
4\\
2
\end{array}\right)(x^{2})^{2}y^{1}=6x^{4}y^{2}\qquad $..OK
4th term: $\left(\begin{array}{l}
4\\
3
\end{array}\right)(x^{2})^{1}y^{3}=4x^{2}y^{3}\qquad $..OK
expression = $(x^{2}+y)^{4} $
