Answer
$x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$
Work Step by Step
Use the binomial theorem formula (detailed on page 754) to solve the problem. Be careful in writing each term correctly.
$(x-3y)^5$
$1*(x)^5+5*(x)^4*(-3y)+10*(x)^3*(-3y)^2+10*(x)^2*(-3y)^3+5*(x)*(-3y)^4+1*(-3y)^5$
$x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$
