Answer
$y^4-12y^3+54y^2-108y+81$
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:
$(y-3)^4=\displaystyle \binom{4}{0}(y)^4(-3)^0+\displaystyle \binom{4}{1}(y)^3(-3)^1+\displaystyle \binom{4}{2}(y)^2(-3)^2+\displaystyle \binom{4}{3}(y)^1(-3)^3+\displaystyle \binom{4}{4}(y)^0(-3)^4$
or, $=y^4+4(y^3)(-3)+6(y^2)(9)+4(y)(-27)+81$
or, $=y^4-12y^3+54y^2-108y+81$