Step 1 :

After factoring out $ 3x $ we have:

$$ 9x^{4}-15x^{3}-9x^{2}+15x = 3x ( 3x^{3}-5x^{2}-3x+5 ) $$

Step 2 :

To factor $ 3x^{3}-5x^{2}-3x+5 $ we can use factoring by grouping:

Group $ \color{blue}{ 3x^{3} }$ with $ \color{blue}{ -5x^{2} }$ and $ \color{red}{ -3x }$ with $ \color{red}{ 5 }$ then factor each group.

$$ \begin{aligned} 3x^{3}-5x^{2}-3x+5 = ( \color{blue}{ 3x^{3}-5x^{2} } ) + ( \color{red}{ -3x+5 }) &= \\ &= \color{blue}{ x^{2}( 3x-5 )} + \color{red}{ -1( 3x-5 ) } = \\ &= (x^{2}-1)(3x-5) \end{aligned} $$

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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