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$(a+b)(a+b+c)=15$
$(b+c)(a+b+c)=6$
$(c+a)(a+b+c)=-3$

$\{ a,b,c\} \equiv ?$

Let $a+b+c=k$. So,

$k(a+b)=15$
$k(b+c)=6$
$k(c+a)=-3$

Adding all of the above, $k.2k=18$

$\Rightarrow k=\pm 3$

Case I: $k=3$

$a+b=\frac {15}{k}=5$
$b+c=\frac {6}{k}=2$
$c+a=\frac {-3}{k}=-1$

$c=(a+b+c)-(a+b)=k-(a+b)=3-5=-2$
$a=(a+b+c)-(b+c)=k-(b+c)=3-2=1$
$b=(a+b+c)-(c+a)=k-(c+a)=3-(-1)=4$

$\therefore \{ a,b,c\}  \equiv \{ 1,4, - 2\} $

Case II: $k=-3$

$a+b=\frac {15}{k}=-5$
$b+c=\frac {6}{k}=-2$
$c+a=\frac {-3}{k}=1$

$c=(a+b+c)-(a+b)=k-(a+b)=-3-(-5)=2$
$a=(a+b+c)-(b+c)=k-(b+c)=-3-(-2)=-1$
$b=(a+b+c)-(c+a)=k-(c+a)=-3-1=-4$

$\therefore \{ a,b,c\}  \equiv \{ -1,-4, 2\} $

Thus, $\{ a,b,c\}  \equiv \{ 1,4, - 2\}  \cup \{  - 1, - 4,2\} $

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