How do you factor #3x^4-81xy^3#?

1 Answer
Jun 22, 2018

Answer:

#3x(x-3y)(x^2+3xy+9y^2)#

Explanation:

#"take out a "color(blue)"common factor "3x#

#=3x(x^3-27y^3)#

#x^3-27y^3" is a "color(blue)"difference of cubes"#

#"which factors in general as"#

#a^3-b^3=(a-b)(a^2+ab+b^2)#

#"here "a=x" and "27y^3=(3y)^3rArrb=3y#

#x^3-27y^3=(x-3y)(x^2+3xy+9y^2)#

#3x^4-81xy^3=3x(x-3y)(x^2+3xy+9y^2)#