# Solve the following equation in the real number system. Please show all of your work. x^3-x^2+8x-8=0calculus

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### 3 Answers

`x^3 - x^2 + 8x -8 = 0`

First we will group the sides to be factored together:

`(x^3 -x^2) + (8x-8) = 0`

Now we will factor x^2 from the first 2 terms and 8 from the last two terms:

`x^2 (x-1) + 8(x-1) = 0`

Now we will factor (x-1):

`(x-1)(x^2 + 8) = 0`

Now we will find the roots (zeros)

`==> x_1 = 1`

`==> x_2 = 2sqrt2*i`

`==> x3= -2sqrt2*i`

**Then, we have one real solution and two complex solutions.**

x^3-x^2+8x-8=0

(x^3-x^2)+(8x-8)

x^2 (x - 1) + 8 ( x -1)

(x^2 + 8) (x - 1)

x - 1 = 0

*x = 1*

x^2 + 8 = 0

x^2 = -8

x^3-x^2+8x-8=0

group:

(x^3-x^2)+(8x-8)

factor out:

x^2 (x - 1) + 8 ( x -1)

(x^2 + 8) (x - 1)

set them equal to 0

x - 1 = 0

*x = 1*

x^2 + 8 = 0

x^2 = -8

`sqrt(x^2) =sqrt(-8)`

`x = +-sqrt(-8)`