Factoring question.

Question

(x-1)^3—8

 

i tried solving but ended up getting (x+1)^2 but its wrong

Answer 1

First, lets expand this.

 

 

(x - 1)(x - 1)(x - 1) - 8 =

 

(x - 1)(x² - 2x + 1) - 8 =

 

x(x² - 2x + 1) - (x² - 2x + 1) - 8 =

 

x³ - 2x² + x - x² + 2x - 1 - 8 =

 

x³ - 3x² + 3x - 9

 

 

Now we can factor by grouping.

 

x²(x - 3) + 3(x - 3)

 

 

(x - 3)(x² + 3)

Answer 2

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Answer 3

(x-1)^3 - 8 = (x-1)^3 - 2^3

 

Note this is a difference of cubes:

 

a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a = (x-1) and b=2

 

Then

 

(x-1)^3 - 2^3 = [(x - 1) - 2] [(x - 1)^2 + (x - 1)(2) + 4]

 

= (x - 3) [(x^2 - 2x + 1) + (2x - 2) + 4]

 

= (x-3) (x^2 + 3), which cannot be further factored unless you allow complex roots.