Answer
$x^{10}-20x^9y+180x^8y^2$
Work Step by Step
Apply Binomial Theorem or Binomial expansion to find the first three terms of $(x-2y)^{10}$.
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
Now, $(x-2y)^{10}=\displaystyle \binom{10}{0}x^{10}(-2y)^0+\displaystyle \binom{10}{1}x^{9}(-2y)^1+\displaystyle \binom{10}{2}x^{8}(-2y)^2$
$=(1)x^{10}(1)+10x^9(-2y)+45x^8(4y^2)$
Thus, $(x-2y)^{10}=x^{10}-20x^9y+180x^8y^2$
