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