Step 1 :

To factor $ 4x^{3}-32x^{2}-81x+648 $ we can use factoring by grouping:

Group $ \color{blue}{ 4x^{3} }$ with $ \color{blue}{ -32x^{2} }$ and $ \color{red}{ -81x }$ with $ \color{red}{ 648 }$ then factor each group.

$$ \begin{aligned} 4x^{3}-32x^{2}-81x+648 = ( \color{blue}{ 4x^{3}-32x^{2} } ) + ( \color{red}{ -81x+648 }) &= \\ &= \color{blue}{ 4x^{2}( x-8 )} + \color{red}{ -81( x-8 ) } = \\ &= (4x^{2}-81)(x-8) \end{aligned} $$

Step 2 :

Rewrite $ 4x^{2}-81 $ as:

$$ 4x^{2}-81 = (2x)^2 - (9)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = 2x $ and $ II = 9 $ , we have:

$$ 4x^{2}-81 = (2x)^2 - (9)^2 = ( 2x-9 ) ( 2x+9 ) $$

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