Answer
$x^3+9x^2y+27xy^2+27y^3$
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:
$(x+3y)^3=\displaystyle \binom{3}{0}(x)^3(3y)^0+\displaystyle \binom{3}{1}(x^{2})(3y)^1
+\displaystyle \binom{3}{2}(x)^1(3y)^2+\displaystyle \binom{3}{3}(x)^0(3y)^3$
or, $=x^3+3(x^2)3y+3(x)(9y^2)+27y^3$
or, $=x^3+9x^2y+27xy^2+27y^3$
