quadratic cocassi

Using Vieta's Formulas, we have:

$A+B=-1\quad\quad AB = 1\quad \quad C+D = -3 \quad \quad CD=1$

$(A-C)(B+D)(A+D)(B-C) = (AB+AD-BC-CD)(AB-AC+BD-CD)$

$=(1+AD-BC-1)(1-AC+BD-1) = (AD-BC)(BD-AC)$

$=ABD^2-A^2CD-B^2CD+ABC^2=AB(D^2+C^2)-CD(A^2+B^2)$

$=(C^2+D^2)-(A^2+B^2) = [(C+D)^2-2CD]-[(A+B)^2-2AB]$

$=9-2-1+2=\boxed{8}$

Let $P(x)=x^2+x+1=(x-A)(x-B)$ and $Q(x)=x^2+3x+1=(x-C)(x-D)$ .

Now, rearrange the expression we want to find: $(C-A)(C-B)(-D-A)(-D-B)$ , which is simply $P(C)P(-D)$ :

$(C^2+C+1)(D^2-D+1)$ .

Expanding we get:

$1+C^2+D^2+C-D+(CD)^2+CD^2-C^2D-CD$

Simplifying it we get:

$1+(C+D)^2-3CD+(CD)^2-(C-D)(1-CD)$

By Vieta's formula, we know that $C+D=-3$ and $CD=1$ , so:

$1+(-3)^2-3(1)+(1)^2-(C-D)(1-1)$

$1+9-3+1=\boxed{8}$

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