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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.

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HINT: Let $p(x)$ be the polynomial. Note that

$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)=(7-d)(7-c)(7-b)(7-a)=p(?)\;.$$

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

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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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