Answer
$x^8+4x^6y+6x^4y^2+4x^2y^3+y^4$
Work Step by Step
According to the Binomial Theorem or Binomial expansion, the expansion for $(x+y)$ in the any number of power can be calculated as:
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
Let's see how this formula works:
$(x^2+y)^4=\displaystyle \binom{4}{0}(x^2)^4(y)^0+\displaystyle \binom{4}{1}(x^2)^3(y)^1+\displaystyle \binom{4}{2}(x^2)^2(y)^2+\displaystyle \binom{4}{3}(x^2)^1(y)^3+\displaystyle \binom{4}{4}(x^2)^0(y)^4$
or, $=x^8+4(x^6)(y)+6(x^4)(y^2)+4x^2y^3+y^4$
or, $=x^8+4x^6y+6x^4y^2+4x^2y^3+y^4$