How do you factor: #y= 27x^3 - 1 #?

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Answer 1 · 2016-01-18T18:01:18

#(27x^3 - 1) = (3x - 1) (9x^2 + 3x + 1)#

Cube root of 27 is 3 and the cube root of -1 is also -1.
#27x^3 - 1 = (3x)^3 - (1)^3#

To factor #y = 27x^3 - 1#,
remember that #(a^3-b^3) = (a-b)(a^2+ab+b^2)#

substituting,

#(27x^3 - 1) = (3x - 1) ((3x)^2 + (3x)(1) + (1)^2)#
#(27x^3 - 1) = (3x - 1) (9x^2 + 3x + 1)#

Answer 2 · 2016-01-18T18:41:05

#(3x-1)(9x^2+3x+1)#

Notice that both the terms are cubed terms, i.e. #27x^3=(3x)^3# and #1=1^3#.

That means that this is a difference of cubes, which can be factored as follows;

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

Thus, we have #a=3x# and #b=1#:

#27x^3-1=(3x)^3-1^3#

#=(3x-1)((3x)^2+(3x)(1)+1^2)#

#=(3x-1)(9x^2+3x+1)#