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.

Let me try. You have $$(a+b+c)(a+b+d)(a+c+d)(b+c+d) = (7-a)(7-b)(7-c)(7-d) = 7^4 - 7^3(a+b+c+d) + 7^2(ab+ac+ad+bc+bd+cd)-7(abc+abd+bcd+acd) + abcd$$
