Answer
$y^{60}-20y^{57}+190x^{54}$
Work Step by Step
Apply Binomial Theorem or Binomial expansion to find the first three terms of $(y^3-1)^{20}$.
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
Now, $(y^3-1)^{20}=\displaystyle \binom{20}{0}(y^3)^{20}(-1)^0+\displaystyle \binom{20}{1}(y^3)^{19}(-1)^1+\displaystyle \binom{20}{2}(y^3)^{18}(-1)^2$
Thus, $(y^3-1)^{20}=y^{60}-20y^{57}+190x^{54}$
