2026/064) Given a,b,c,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$ Evaluate $(a+b+c)(b+c+d)(c+d+a)(d+a+b)$

Because a,b ,c ,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$

We have by vieta's formula
 

$a+b+c+d = 7\cdots(1)$
 

Let
 

$f(x) = x^4−7x^3+3x^2−21x+1=(x-a)(x-b)(x-c)(x-d)\cdots(2)$
 

From (1) we have
 

$a+b+c = 7-d\cdots(3)$
 

$b+c+d  = 7 -a\cdots(4)$
 

$c+d+a = 7 -b\cdots(5)$
 

 $d+a +b = 7-c\cdots(6)$
 

From (3), (4), (5),(6) we have
 

$(a+b+c)(b+c+d)(c+d+a)(d+a+b)= (7-d)(7-a)(7- b)(7-c) = f(7)\cdots(7)$ from (2)
 

As $f(x) = x^4−7x^3+3x^2−21x+1$
 

So $f(7) = 7^4 - 7 * 7^3 +3 *7^2 - 21 * 7 + 1 = 1$
 

From (7) and above we have
 

$(a+b+c)(b+c+d)(c+d+a)(d+a+b) = 1$