# How to solve this equation x^3+3x^2-x-1=0 with such substitution as x=y-1?

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You should substitute y-1 for x in equation above such that:

`(y - 1)^3 + 3(y - 1)^2 - y + 1 - 1 = 0`

`(y - 1)^3 + 3(y - 1)^2 - y = 0`

`y^3 - 3y^2 + 3y - 1 + 3y^2 - 6y + 3 - y = 0`

`y^3 - 4y + 2 = 0`

Notice that sketching the graph of the function yields that there exists three intersections to x axis, hence, the equation has three solutions.

You should come up with the following notations for these solutions such that:

`y_1 in (-2.5, -2), y_2 in (0, 1), y_3 in (1, 2)`

**You should come back to the substitution `x = y-1` , hence, the solutions to the given equation are: `x_1 in (-3.5, -3), x_2 in (-1, 0), x_3 in (0,1).` **

but i need exact answers....and i need to solve it with using no graphs