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

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`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.

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