Question
$$(x+2)^3-6(x+2)^2+11(x+2)-6 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = -1 \end{matrix} $$
Explanation
$$ \begin{aligned} (x+2)^3-6(x+2)^2+11(x+2)-6 &= 0&& \text{simplify left side} \\[1 em]x^3+6x^2+12x+8-6(x^2+4x+4)+11(x+2)-6 &= 0&& \\[1 em]x^3+6x^2+12x+8-(6x^2+24x+24)+11x+22-6 &= 0&& \\[1 em]x^3+6x^2+12x+8-6x^2-24x-24+11x+22-6 &= 0&& \\[1 em]x^3-12x-16+11x+22-6 &= 0&& \\[1 em]x^3-x+6-6 &= 0&& \\[1 em]x^3-x+6-6 &= 0&& \\[1 em]x^3-x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}-x = 0 } $, first we need to factor our $ x $.
$$ x^{3}-x = x \left( x^{2}-1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{2}-1 = 0$.
$ x^{2}-1 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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