Answer
$(x+y)^{4}$
Work Step by Step
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
-------------------
The coefficients are 1, 4, 6, 4, 1
$\left(\begin{array}{l}
4\\
0
\end{array}\right)=1=\left(\begin{array}{l}
4\\
4
\end{array}\right)$
$\left(\begin{array}{l}
4\\
1
\end{array}\right)=4=\left(\begin{array}{l}
4\\
3
\end{array}\right)$
$\displaystyle \left(\begin{array}{l}
4\\
2
\end{array}\right)=\frac{4\times 3}{1\times 2}=6$
The exponents of x decrease from 4 to zero
The exponents of y increase from 0 to 4.
$x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}=(x+y)^{4}$