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

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$