How do you factor #27x^3 - 27y^3#?

2 Answers
Oct 4, 2016

Answer:

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

Explanation:

This is the difference of cubes and has a set pattern.

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

make 2 brackets:
The first is simply the cube root of each term

The second bracket is formed from the first in 5 steps:

#1.rarr# square the first term.
#2.rarr# change the sign
#3.rarr# multiply the two terms
#4.rarr# PLUS
#5.rarr#square the second term

Oct 4, 2016

Answer:

#27(x-y)(x^2+xy+y^2)#

Explanation:

The first step here is to take out a #color(blue)"common factor"# of 27.

#rArr27(x^3-y^3)........ (A)#

Now #x^3-y^3# is a #color(blue)"difference of cubes"# and in general factorises as.

#color(red)(bar(ul(|color(white)(a/a)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(a/a)|)))#

here a = x and b = y

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

Substituting this result into (A) gives the factorised form.

#rArr27x^3-27y^3=27(x-y)(x^2+xy+y^2)#