$$(x+2)\frac{x-2}{x+1}(x-1)\frac{x+1}{x}(x+2) = 0$$
Answer
Explanation
In order to solve $ \color{blue}{ x^{6}+x^{5}-6x^{4}-4x^{3}+8x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ x^{6}+x^{5}-6x^{4}-4x^{3}+8x^{2} = x^2 \left( x^{4}+x^{3}-6x^{2}-4x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ x^{4}+x^{3}-6x^{2}-4x+8 = 0$.
$ \color{blue}{ x^{4}+x^{3}-6x^{2}-4x+8 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.
The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.
The factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (8) are 1 2 4 8 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } ~ $$Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
If we plug these values into the polynomial $ P(x) $, we obtain $ P(1) = 0 $.
To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:
Divide $ P(x) $ with $ \color{blue}{x - 1} $
$$ \frac{ x^{4}+x^{3}-6x^{2}-4x+8 }{ \color{blue}{ x - 1 } } = x^{3}+2x^{2}-4x-8 $$Polynomial $ x^{3}+2x^{2}-4x-8 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+2x^{2}-4x-8 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.