<s>[INST] How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial: 
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to apply Vieta formula to find the relationship between its roots. From there, I can get:
$$a+b+c+d = 7$$
$$ab+ac+ad+bc+bd+cd= 3$$
$$abc+abd+acd+bcd= 21$$
$$abcd= 1$$
Then,
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$$
$$= (a²+b²+ab+3)(c²+d²+cd+3)$$
$$= 3(a^2+b^2+c^2+d^2)+abcd+(ac)^2+(ad)^2+(bc)^2+(bd)^2+a^2cd+b^2cd+abc^2+abd^2$$
From there, I don't have any idea how to go further.
Can somebody help me to explain to solve this equation?
Thanks.
 [/INST] Based on Vieta's formula and Brian and GAVD answers:
From Vieta formula, we got:
$a+b+c+d=7$
$ab+ac+ad+bc+bd+cd=3$
$abc+abd+acd+bcd=21$
$abcd=1$
Then:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$= (a+b+c+d-d)(a+b+c+d-c)(a+b+c+d-b)(a+b+c+d-a)$
$= (7-d)(7-c)(7-b)(7-a)$
Simplify and we got:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$=(7−a)(7−b)(7−c)(7−d)$
$=7^4−7^3(a+b+c+d)+7^2(ab+ac+ad+bc+bd+cd)\\ \ \ −7(abc+abd+bcd+acd)+abcd$
Substitute all values and we got:
$=7^4-7^3(7)+7^2(3)-7(21)+1$
$=7(21)-7(21)+1$
$=1$
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