Step 1 :

To factor $ 3x^2-3+x^2y-y $ we can use factoring by grouping.

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

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

Step 2 :

Rewrite $ x^2-1 $ as:

$$ \color{blue}{ x^2-1 = (x)^2 - (1)^2 } $$

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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