Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1092: 28

$x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$

Based on the Binomial Theorem, we have $(x-3y)^5=\binom50 x^5(-3y)^0+\binom51 x^4(-3y)^1+ \binom52 x^3(-3y)^2+\binom53 x^2(-3y)^3+\binom54 x^1(-3y)^4+\binom55 x^0(-3y)^5=x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$