Factor polynomial $$ x^{6}-x^{5}-x^{4}+x^{3} $$

$$ x^{6}-x^{5}-x^{4}+x^{3} = x^{3}( x-1 )^{ 2 }( x+1 ) $$

Step 1 :

After factoring out $ x^{3} $ we have:

$$ x^{6}-x^{5}-x^{4}+x^{3} = x^{3} ( x^{3}-x^{2}-x+1 ) $$

Step 2 :

To factor $ x^{3}-x^{2}-x+1 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ -x^{2} }$ and $ \color{red}{ -x }$ with $ \color{red}{ 1 }$ then factor each group.

$$ \begin{aligned} x^{3}-x^{2}-x+1 = ( \color{blue}{ x^{3}-x^{2} } ) + ( \color{red}{ -x+1 }) &= \\ &= \color{blue}{ x^{2}( x-1 )} + \color{red}{ -1( x-1 ) } = \\ &= (x^{2}-1)(x-1) \end{aligned} $$

Step 3 :

Rewrite $ x^{2}-1 $ as:

$$ x^{2}-1 = (x)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x $ and $ II = 1 $ , we have:

$$ x^{2}-1 = (x)^2 - (1)^2 = ( x-1 ) ( x+1 ) $$

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