How do you find the product #(4y+3z)(4y-3z)^2#?

1 Answer
Mar 25, 2017

Answer:

#(4y+3z)(4y-3z)^2=color(green)(64y^3-48y^2z-36yz^2+27z^3)#

Explanation:

Using the difference of squares we could write
#(4y+3z)(4y-3z)^2=color(blue)((4y+3z)(4y-3z))(4y-3z)#

#color(white)("XXXXXXXXXXX")=color(blue)(((4y)^2-(3z)^2))(4y-3z)#

#color(white)("XXXXXXXXXXX")=(16y^2-9z^2)(4y-3)#

Then using the distributive property or tabular multiplication:
#{:(underline(xx)," | ",underline(+16y^2),underline(-9z^2)), (+4y," | ",+64y^3,-36yz^2), (underline(-3z),ul(" | "),ul(-48y^2z),ul(+27z^3)), (color(green)(+64y^3),color(green)(-48y^2z),color(green)(-36yz^2),color(green)(+27z^3)) :}#