Okay, so I need to find the set of sums of products of roots taken two at a time for a quartic equation. The equation given is x⁴ + 2x³ + 2 = 0, and the roots are α₁, α₂, α₃, α₄. The task is to determine the unordered set {α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃}. Hmm, let me start by recalling some properties of polynomials and their roots.

First, for a quartic equation x⁴ + a x³ + b x² + c x + d = 0, the sums of roots taken one at a time is -a, the sum of products two at a time is b, three at a time is -c, and the product of all four roots is d. But in this case, the given equation is x⁴ + 2x³ + 2 = 0, so comparing to the general form, a = 2, b = 0 (since there's no x² term), c = 0 (no x term), and d = 2. Wait, is that right? Let me check: the equation is x⁴ + 2x³ + 0x² + 0x + 2 = 0. So, yes, coefficients are a=2, b=0, c=0, d=2.

So applying Vieta's formula:

Sum of roots: α₁ + α₂ + α₃ + α₄ = -a = -2.

Sum of products two at a time: α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = b = 0.

Sum of products three at a time: α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -c = 0.

Product of roots: α₁α₂α₃α₄ = d = 2.

But the question is about the sums α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃. These are the three possible ways to partition the four roots into two pairs and sum the products of each pair. So, we need to find these three values.

Let me denote S₁ = α₁α₂ + α₃α₄, S₂ = α₁α₃ + α₂α₄, S₃ = α₁α₄ + α₂α₃. The problem is to find the set {S₁, S₂, S₃}.

Since all these S_i are symmetric in the roots, perhaps there's a way to relate them to the coefficients of the original polynomial. Let me think. The sum S₁ + S₂ + S₃ would be equal to the sum of all possible products of two roots, which is exactly the second symmetric sum, which we know is 0. So, S₁ + S₂ + S₃ = 0. That's useful.

But we need each individual S_i. How can we find them? Maybe we need to find more equations involving these S_i. Let me consider the square of the sum of the roots:

(α₁ + α₂ + α₃ + α₄)² = α₁² + α₂² + α₃² + α₄² + 2(α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄). We know the sum of the roots is -2, so (-2)² = 4 = sum of squares + 2*(sum of products). The sum of products is 0, so 4 = sum of squares. Therefore, the sum of squares of the roots is 4. Not sure if that helps directly, but maybe.

Alternatively, perhaps we can consider the products of the roots taken three at a time. Wait, the sum of products three at a time is zero. Let me think. Each term in that sum is a product of three roots. For example, α₁α₂α₃ + α₁α₂α₄ + ... etc. If we multiply each of these by the remaining root, which is the fourth root, we get the product of all four roots, which is 2. So, for each triplet product, multiplying by the remaining root gives the product of all four roots. So, α₁α₂α₃ * α₄ = α₁α₂α₃α₄ = 2, similarly for others. Therefore, α₁α₂α₃ = 2 / α₄, and so on.

But how does that help with the S_i? Let's see. Let's take S₁ = α₁α₂ + α₃α₄. If I multiply S₁ by (α₁ + α₂ + α₃ + α₄), which is -2, we get:

S₁*(-2) = α₁α₂(α₁ + α₂ + α₃ + α₄) + α₃α₄(α₁ + α₂ + α₃ + α₄).

Expanding the first term: α₁²α₂ + α₁α₂² + α₁α₂α₃ + α₁α₂α₄.

Similarly, the second term: α₁α₃α₄ + α₂α₃α₄ + α₃²α₄ + α₃α₄².

But this seems complicated. Maybe there's a better approach. Let's think about the original polynomial. Since the roots satisfy x⁴ + 2x³ + 2 = 0, each root α satisfies α⁴ = -2α³ - 2. Maybe we can use this to express higher powers of α in terms of lower powers.

Alternatively, consider that the polynomial can be factored into quadratics. Let me check if it's possible to factor x⁴ + 2x³ + 2 into two quadratics. Suppose (x² + a x + b)(x² + c x + d) = x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + bd. Comparing coefficients:

a + c = 2

ac + b + d = 0

ad + bc = 0

bd = 2

So we have a system of equations. Let's try to solve this. Since bd = 2, possible integer pairs for (b, d) are (1,2), (2,1), (-1,-2), (-2,-1). Let's try b=1, d=2. Then:

a + c = 2

ac + 1 + 2 = ac + 3 = 0 => ac = -3

ad + bc = a*2 + c*1 = 2a + c = 0

From a + c = 2, c = 2 - a. Substitute into 2a + c = 0:

2a + (2 - a) = a + 2 = 0 => a = -2. Then c = 2 - (-2) = 4.

Check ac = (-2)(4) = -8 ≠ -3. Doesn't work.

Next, try b=2, d=1:

ac + 2 + 1 = ac + 3 = 0 => ac = -3

ad + bc = a*1 + c*2 = a + 2c = 0

From a + c = 2, a = 2 - c. Substitute into a + 2c = 0:

(2 - c) + 2c = 2 + c = 0 => c = -2. Then a = 2 - (-2) = 4.

Check ac = 4*(-2) = -8 ≠ -3. Not good.

Try b=-1, d=-2:

ac + (-1) + (-2) = ac -3 = 0 => ac = 3

ad + bc = a*(-2) + c*(-1) = -2a - c = 0

From a + c = 2, c = 2 - a. Substitute into -2a - c = 0:

-2a - (2 - a) = -2a -2 + a = -a -2 = 0 => -a = 2 => a = -2. Then c = 2 - (-2) = 4.

Check ac = (-2)(4) = -8 ≠ 3. Nope.

Next, b=-2, d=-1:

ac + (-2) + (-1) = ac -3 = 0 => ac = 3

ad + bc = a*(-1) + c*(-2) = -a -2c = 0

From a + c = 2, a = 2 - c. Substitute into -a -2c = 0:

-(2 - c) -2c = -2 + c -2c = -2 - c = 0 => -c = 2 => c = -2. Then a = 2 - (-2) = 4.

Check ac = 4*(-2) = -8 ≠ 3. Nope.

So none of the integer pairs work. Maybe the polynomial is irreducible over integers. Therefore, factoring into quadratics with integer coefficients isn't possible. Maybe with real coefficients? Hmm, but solving a quartic might be complex. Alternatively, perhaps using symmetric sums and Newton's identities?

Wait, Newton's identities relate power sums to elementary symmetric sums. The power sums are p_k = α₁^k + α₂^k + α₃^k + α₄^k. The elementary symmetric sums are the coefficients we already have: s1 = -2, s2 = 0, s3 = 0, s4 = 2.

Newton's identities give a way to compute p_k in terms of s1, s2, s3, s4. Let's see if we can use this.

First, we know p1 = α₁ + α₂ + α₃ + α₄ = -2.

p2 = α₁² + α₂² + α₃² + α₄². We can compute this using p1² - 2s2. Wait, s2 is the sum of products two at a time, which is 0. So p2 = (-2)^2 - 2*0 = 4.

Similarly, p3 = α₁³ + α₂³ + α₃³ + α₄³. Newton's formula says p3 = s1 p2 - s2 p1 + 3 s3. Plugging in the values: (-2)(4) - 0*(-2) + 3*0 = -8 + 0 + 0 = -8.

p4 = α₁⁴ + α₂⁴ + α₃⁴ + α₄⁴. Using the original equation, each α_i satisfies α_i⁴ = -2α_i³ - 2. Therefore, α_i⁴ = -2α_i³ - 2. So sum over all roots:

p4 = sum α_i⁴ = -2 sum α_i³ - 2*4 = -2*(-8) -8 = 16 -8 = 8.

Wait, let me check that again. Each α_i⁴ = -2α_i³ -2, so summing over all four roots:

sum α_i⁴ = -2 sum α_i³ - 2*4. Because each term is -2α_i³ -2, so summing four times gives -2*(sum α_i³) - 8. Yes, so sum α_i⁴ = -2*(-8) -8 = 16 -8 = 8. So p4 = 8.

Now, using Newton's identities, we can compute p4 in terms of s1, s2, s3, s4 and previous p's.

Newton's formula for p4 is s1 p3 - s2 p2 + s3 p1 - s4 p0, where p0 = 4 (since p0 = number of variables). Wait, let me check the exact formula. Newton's identities for quartic:

p4 = s1 p3 - s2 p2 + s3 p1 - s4 p0

Given that p0 = 4 (the number of variables, since there are four roots), and s4 = 2 (the product of all roots). So:

p4 = (-2)(-8) - 0*4 + 0*(-2) - 2*4 = 16 - 0 + 0 -8 = 8. Which matches our previous result. So that's correct.

But how does this help us find S₁, S₂, S₃? Hmm. Maybe we need another approach. Let me think.

Alternatively, consider that the products S₁, S₂, S₃ satisfy a certain cubic equation. Since the original polynomial is quartic, and the products S₁, S₂, S₃ might be roots of a cubic resolvent. Maybe we can find such a resolvent cubic.

The general method for solving quartic equations involves finding a resolvent cubic whose roots are precisely these S₁, S₂, S₃. If we can construct that cubic, then solving it would give us the values of S₁, S₂, S₃.

So let's try to construct the resolvent cubic. Let me recall that for a quartic equation x⁴ + a x³ + b x² + c x + d = 0, the resolvent cubic is t³ - b t² + (ac - 4d) t - (a² d - 4b d + c²) = 0. Wait, I'm not sure if I remember this correctly. Maybe I need to check the formula.

Alternatively, the resolvent cubic can be constructed using the products of roots taken two at a time. Wait, but S₁, S₂, S₃ are sums of products of roots taken two at a time. Let me check.

Let me denote the roots as α, β, γ, δ. Then the resolvent cubic is (x - (αβ + γδ))(x - (αγ + βδ))(x - (αδ + βγ)). This is exactly the cubic whose roots are the S_i's. So to find this cubic, we need to compute its coefficients.

Let me denote the cubic as x³ - p x² + q x - r = 0. Then:

p = S₁ + S₂ + S₃ = 0 (from earlier, since sum of products two at a time is 0).

q = S₁ S₂ + S₁ S₃ + S₂ S₃.

r = S₁ S₂ S₃.

So we need to compute q and r.

First, p is already known to be 0. So the cubic is x³ + q x - r = 0.

To find q and r, we need expressions in terms of the coefficients of the original quartic.

I recall that for quartic equations, the coefficients of the resolvent cubic can be expressed in terms of the original coefficients. Let me try to derive them.

Given the quartic equation x⁴ + a x³ + b x² + c x + d = 0, the resolvent cubic is:

t³ - b t² + (a c - 4 d) t - (a² d - 4 b d + c²) = 0.

Wait, let's check this. If that's the case, then for our quartic x⁴ + 2x³ + 2 = 0, a = 2, b = 0, c = 0, d = 2. Then substituting into the resolvent cubic formula:

t³ - 0 t² + (2*0 - 4*2) t - (2²*2 - 4*0*2 + 0²) = t³ - 8 t - (8 - 0 + 0) = t³ - 8 t -8.

Therefore, the resolvent cubic would be t³ -8 t -8 = 0. Then the roots of this cubic are S₁, S₂, S₃. Therefore, the set {S₁, S₂, S₃} is the set of roots of the equation t³ -8t -8 = 0.

But let me verify if this formula is correct. The resolvent cubic for a quartic is indeed given by t³ - b t² + (a c - 4 d) t - (a² d - 4 b d + c²). Let's confirm with a source. Wait, since I can't look it up, let me try to derive it.

Alternatively, let's compute q and r.

q = S₁ S₂ + S₁ S₃ + S₂ S₃.

r = S₁ S₂ S₃.

We need expressions for q and r in terms of the original coefficients.

Let me compute q first.

Compute S₁ S₂ + S₁ S₃ + S₂ S₃.

First, expand S₁ S₂:

(α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄) = α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄².

Similarly, S₁ S₃ = (α₁α₂ + α₃α₄)(α₁α₄ + α₂α₃) = α₁²α₂α₄ + α₁α₂²α₃ + α₁α₄²α₃ + α₂α₃²α₄.

And S₂ S₃ = (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃) = α₁²α₃α₄ + α₁α₃²α₄ + α₁α₂α₃² + α₂²α₃α₄.

Adding all these together:

Sum = [α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²] + [α₁²α₂α₄ + α₁α₂²α₃ + α₁α₄²α₃ + α₂α₃²α₄] + [α₁²α₃α₄ + α₁α₃²α₄ + α₁α₂α₃² + α₂²α₃α₄].

Combine like terms:

- Terms with α₁²α₂α₃: 1
- Terms with α₁²α₂α₄: 1
- Terms with α₁²α₃α₄: 1
- Terms with α₂²α₃α₄: 1
- Terms with α₂²α₃α₄: 1 (from S₂ S₃)
- Terms with α₁α₂²α₄: 1
- Terms with α₁α₂²α₃: 1
- Terms with α₁α₃²α₄: 2 (from S₁ S₂ and S₁ S₃)
- Terms with α₁α₄²α₃: 1
- Terms with α₂α₃²α₄: 2 (from S₁ S₂ and S₂ S₃)
- Terms with α₂α₄²α₃: 1
- Terms with α₁α₂α₃²: 1 (from S₂ S₃)
- Terms with α₂²α₃α₄: 1 (from S₂ S₃)

Wait, this is getting complicated. Maybe there's a better way. Let me consider the sum S₁ S₂ + S₁ S₃ + S₂ S₃. Let's denote this sum as Q. Then:

Q = sum_{i<j, k<l, (i,j,k,l) distinct} (α_i α_j)(α_k α_l).

But since we're summing over all possible products of two pairs that don't overlap, but since the four roots are partitioned into two pairs, each term is a product of two disjoint pairs. However, the sum over all such products is equal to (sum α_i α_j)^2 - 2 sum α_i α_j α_k α_l, where i < j < k < l. Wait, maybe.

Wait, the square of the sum of all possible two-element products is equal to the sum of squares of two-element products plus twice the sum of products of two-element products multiplied by another two-element products. But perhaps it's more precise to say:

(sum_{i<j} α_i α_j)^2 = sum_{i<j} (α_i α_j)^2 + 2 sum_{i<j, k<l, (i,j) ≠ (k,l)} (α_i α_j)(α_k α_l).

But we want the sum over all distinct pairs of two-element products. Wait, but in our case, the two-element products are grouped into three different pairings. Each of S₁, S₂, S₃ are such pairings. However, the square of the sum of all two-element products is equal to the sum of squares of each two-element product plus twice the sum of all possible products of two-element products taken two at a time, where the two products are disjoint. That is:

(sum_{i<j} α_i α_j)^2 = sum_{i<j} (α_i α_j)^2 + 2 sum_{(i,j) < (k,l), disjoint} (α_i α_j)(α_k α_l).

But in our case, the sum Q = S₁ S₂ + S₁ S₃ + S₂ S₃ is exactly the second term on the right-hand side. Since the sum over all possible disjoint pairings is Q, and there are three such pairings (S₁, S₂, S₃). However, the total number of such disjoint pairings for four elements is 3. Therefore, Q is equal to the sum over all three such products.

But the square of the sum of all two-element products is equal to sum_{i<j} (α_i α_j)^2 + 2Q.

So, if we can compute sum_{i<j} (α_i α_j)^2, then we can find Q.

Sum_{i<j} (α_i α_j)^2 is equal to [sum_{i<j} α_i α_j]^2 - 2 sum_{i<j < k < l} α_i α_j α_k α_l. Wait, because (sum α_i α_j)^2 = sum (α_i α_j)^2 + 2 sum_{i<j, k<l, (i,j) ≠ (k,l)} (α_i α_j)(α_k α_l). Then, the sum over all (i,j) ≠ (k,l) is equal to the sum over all possible pairs of two-element products minus the sum over the three disjoint pairings. Wait, this seems confusing.

Alternatively, note that sum_{i<j} (α_i α_j)^2 = (sum α_i)^2 (sum α_j)^2 - 2 sum α_i^2 α_j^2 - 2 (sum α_i)^4 + 2 (sum α_i^2)^2. Wait, no, maybe not. Let me think again.

Actually, sum_{i<j} (α_i α_j)^2 = [sum_{i=1}^4 α_i]^2 [sum_{j=1}^4 α_j]^2 - sum_{i=1}^4 α_i^4 - 2 sum_{i<j} α_i^2 α_j^2. Wait, not sure.

Alternatively, consider that sum_{i<j} (α_i α_j)^2 = [sum_{i=1}^4 α_i^2 sum_{j=1}^4 α_j^2 - sum_{i=1}^4 α_i^4]/2. Because:

sum_{i<j} (α_i α_j)^2 = 1/2 [ (sum α_i^2)^2 - sum α_i^4 ].

Yes, that's a standard identity. Because (sum α_i^2)^2 = sum α_i^4 + 2 sum_{i<j} α_i^2 α_j^2. Therefore, sum_{i<j} (α_i α_j)^2 = [(sum α_i^2)^2 - sum α_i^4]/2.

So, in our case, sum_{i<j} (α_i α_j)^2 = [(sum α_i^2)^2 - sum α_i^4]/2.

We already have sum α_i^2 = 4, from earlier. And sum α_i^4 = p4 = 8. So:

sum_{i<j} (α_i α_j)^2 = [(4)^2 - 8]/2 = (16 -8)/2 = 8/2 = 4.

Therefore, sum_{i<j} (α_i α_j)^2 = 4.

Then, going back to the square of the sum of two-element products:

(sum_{i<j} α_i α_j)^2 = sum (α_i α_j)^2 + 2Q.

We know sum_{i<j} α_i α_j = 0 (from Vieta). Therefore, 0^2 = 4 + 2Q => 0 = 4 + 2Q => Q = -2.

Thus, q = Q = -2.

Now, let's compute r = S₁ S₂ S₃.

To compute r, we need to find the product of S₁, S₂, S₃. Let's denote this as R.

R = S₁ S₂ S₃.

This can be expressed as:

R = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃).

Let me try to expand this. Let's denote A = α₁, B = α₂, C = α₃, D = α₄.

Then R = (AB + CD)(AC + BD)(AD + BC).

Multiply the first two factors:

(AB + CD)(AC + BD) = AB*AC + AB*BD + CD*AC + CD*BD.

= A²BC + AB²D + ACD² + BC D².

Now multiply this by (AD + BC):

[ A²BC + AB²D + ACD² + BC D² ] * (AD + BC)

Let's distribute each term:

First term: A²BC * AD = A³ B C D

Second term: A²BC * BC = A² B² C²

Third term: AB²D * AD = A² B² D²

Fourth term: AB²D * BC = A B³ C D

Fifth term: ACD² * AD = A² C D³

Sixth term: ACD² * BC = A B C² D²

Seventh term: BC D² * AD = A B C D³

Eighth term: BC D² * BC = B² C² D²

So combining all terms:

A³ B C D + A² B² C² + A² B² D² + A B³ C D + A² C D³ + A B C² D² + A B C D³ + B² C² D².

Now, collect like terms:

- Terms with A³ B C D: 1
- Terms with A² B² C²: 1
- Terms with A² B² D²: 1
- Terms with A B³ C D: 1
- Terms with A² C D³: 1
- Terms with A B C² D²: 1
- Terms with A B C D³: 1
- Terms with B² C² D²: 1

So R = A³ B C D + A² B² C² + A² B² D² + A B³ C D + A² C D³ + A B C² D² + A B C D³ + B² C² D².

Now, note that each term can be expressed in terms of the roots. Let me factor out common terms where possible.

First, note that A³ B C D = A² * A B C D. Similarly, A B³ C D = B² * A B C D. Similarly, A² C D³ = C² * A C D². Wait, maybe not. Let me check each term.

But perhaps it's better to compute R in terms of the coefficients of the polynomial. Remember that the roots satisfy α_i^4 = -2α_i³ -2. Let's use this to express higher powers in terms of lower powers.

But this might get complicated. Let me see.

First, note that A³ B C D: Let's compute this term. Since A = α₁, B = α₂, C = α₃, D = α₄.

But in terms of the original equation, each root satisfies α_i^4 = -2α_i³ -2. So, for example, A^4 = -2A³ -2. Similarly for B, C, D.

But A³ B C D = A³ * B * C * D. Let's see if we can express this in terms of other terms.

Alternatively, note that B C D = (B C D). Since the product of all four roots is 2 (from Vieta), B C D = 2 / A. Similarly, A^3 * B C D = A^3 * (2 / A) = 2 A².

Similarly, A B³ C D: A * B³ * C * D = B³ * (A C D). A C D = 2 / B, so this term is B³ * (2 / B) = 2 B².

Similarly, A² B² C²: (A B)^2. Since A B is a product of two roots, and we have other terms.

Wait, maybe this approach is better. Let's compute each term in R:

1. A³ B C D = A² * A B C D = A² * (A B C D). Since A B C D = 2 (product of all roots). Wait, no. Wait, A B C D is the product of all four roots, which is 2 (from Vieta). So A³ B C D = A² * 2.

Similarly:

2. A² B² C² = (A B)^2. We know that A B is a product of two roots. Let me denote AB = m, AC = n, AD = p, etc. But maybe not helpful.

Wait, but from Vieta, we have the sum of products of two roots, which is 0. So sum_{i<j} α_i α_j = 0. The sum of products of two roots is zero. Therefore, the sum of all pairwise products is zero.

But the sum of squares of pairwise products is 4, as we computed earlier.

But perhaps not directly helpful.

Wait, let's go back to term by term.

1. A³ B C D = A² * (A B C D) = A² * 2 (since A B C D = 2). Therefore, 2 A².

2. A² B² C² = (A B)^2. Let me denote this as (A B)^2. Similarly, other terms.

But since we have multiple terms like this, maybe we can express the entire expression in terms of symmetric sums.

Alternatively, note that R is equal to (AB + CD)(AC + BD)(AD + BC). There's a known identity for this product. Wait, I recall that for four variables, (ab + cd)(ac + bd)(ad + bc) = (a + b + c + d)(abc + abd + acd + bcd) - abcd(a + b + c + d). But not sure. Let me check.

Wait, actually, there's an identity that says:

(ab + cd)(ac + bd)(ad + bc) = (a + b)(c + d)(ac + bd) - abcd.

But maybe not useful. Alternatively, perhaps using the relation that (ab + cd)(ac + bd)(ad + bc) = (a + b + c + d)(abc + abd + acd + bcd) - 4abcd.

But let me test this with actual numbers.

Let me take a = 1, b = 2, c = 3, d = 4.

Compute LHS: (1*2 + 3*4)(1*3 + 2*4)(1*4 + 2*3) = (2 +12)(3 +8)(4 +6) = 14 *11 *10 = 1540.

RHS: (1+2+3+4)(1*2*3 +1*2*4 +1*3*4 +2*3*4) - 4*1*2*3*4 = 10*(6 +8 +12 +24) - 96 = 10*50 -96 = 500 -96 = 404. Which is not equal to 1540. So the identity is incorrect. Maybe another one.

Alternatively, perhaps (ab + cd)(ac + bd)(ad + bc) = (a + b + c + d)(ab + ac + ad + bc + bd + cd) - (a^2 b c + a^2 b d + a^2 c d + b^2 a c + b^2 a d + b^2 c d + c^2 a b + c^2 a d + c^2 b d + d^2 a b + d^2 a c + d^2 b c) + 4 a b c d. But this is getting too complicated.

Alternatively, maybe use the fact that (ab + cd)(ac + bd)(ad + bc) = (a + b)(c + d)(ab + cd) - abcd(a + b + c + d). Wait, testing with a=1, b=2, c=3, d=4:

(1+2)(3+4)(1*2 +3*4) -1*2*3*4*(1+2+3+4) = 3*7*14 -24*10 = 294 -240 = 54. But LHS was 1540, so not matching.

Alternatively, maybe I need to look for another identity. Alternatively, use the relation that:

(ab + cd)(ac + bd)(ad + bc) = (a + b)(c + d)(ac + bd) - abcd. But again, testing:

(1+2)(3+4)(1*3 +2*4) -1*2*3*4 = 3*7*11 -24 = 231 -24 = 207. Not equal to 1540.

Hmm. Maybe there's no simple identity here. Therefore, perhaps we need to compute R by expanding it and then express each term in terms of the roots' sums and products.

Wait, but since all terms are products of the roots, perhaps we can relate them to the coefficients. Let's see.

Given that the roots satisfy α_i^4 = -2α_i³ -2, so any higher power can be reduced. For example, α_i^4 = -2α_i³ -2.

Therefore, α_i³ = (-1/2)α_i^4 -1.

But maybe we can use this to express α_i³ in terms of α_i^4. Wait, but in our case, we have terms up to α_i³. Wait, no, in the expansion of R, we have terms like A³, B³, etc. But if we can express A³ in terms of lower powers, but since the equation is quartic, we have α_i^4 = -2α_i³ -2, so α_i³ = (-1/2)α_i^4 -1. Wait, but substituting that into R would complicate things. Maybe not helpful.

Alternatively, note that R is symmetric in the roots, so by the fundamental theorem of symmetric polynomials, R can be expressed in terms of the elementary symmetric sums. Therefore, we can write R in terms of s1, s2, s3, s4.

But since we already know s1 = -2, s2 = 0, s3 = 0, s4 = 2, perhaps we can find R using these.

But let's see. Let's note that R = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃). Let me denote t1 = α₁α₂, t2 = α₁α₃, t3 = α₁α₄, t4 = α₂α₃, t5 = α₂α₄, t6 = α₃α₄.

Then, the three terms are t1 + t5, t2 + t6, t3 + t4.

So R = (t1 + t5)(t2 + t6)(t3 + t4).

But since all these t's are products of two roots, and we know the sum of all t's is zero, and sum of products of two t's is -2, and sum of products of three t's is R.

But Vieta's formula gives us the elementary symmetric sums of the roots, not the t's. However, the sum of t's is the sum of products of two roots, which is zero. The sum of products of two t's is the sum of products of four roots taken two at a time, which is the sum of products of four roots two at a time, which is equal to s2 = 0. Wait, no. The sum of products of two t's (each t is a product of two roots) would be the sum over all pairs of products of two roots, which is the same as the sum of products of four roots taken four at a time? Wait, no.

Wait, more carefully: Each t is a product of two roots. The product of two t's would be a product of four roots. For example, (t1 + t5)(t2 + t6) would include terms like t1 t2, t1 t6, t5 t2, t5 t6. Each t1 t2 is (α₁α₂)(α₁α₂) = α₁² α₂². Wait, but these are products of four roots, but with possible duplicates. Wait, no, t1 t2 = (α₁α₂)(α₁α₂) = α₁² α₂². Similarly, t1 t6 = (α₁α₂)(α₂α₄) = α₁ α₂² α₄. So, the cross terms are products of four roots with some multiplicities.

But this seems messy. Perhaps a better approach is needed.

Alternatively, let's note that R is equal to the product of the three sums S₁, S₂, S₃. But we can also note that the product of these three sums is equal to the product of the three pairings. Since the four roots are being paired in all possible ways, maybe there's a relation.

Alternatively, think of the roots as variables and consider the equations they satisfy. But this might not be helpful.

Alternatively, recall that R is the determinant of a certain matrix, but I don't see an immediate connection.

Alternatively, maybe use the fact that the roots are the roots of the polynomial x⁴ + 2x³ + 2 = 0. Therefore, we can express the product R in terms of the coefficients. But I'm not sure how.

Alternatively, perhaps compute R numerically. But since the polynomial is quartic and the roots are not nice, that might not be feasible.

Wait, but maybe we can use the resolvent cubic. Earlier, we found that the resolvent cubic is t³ -8t -8 = 0. Therefore, if we can compute the roots of this cubic, they will be the S₁, S₂, S₃. Then, we can just solve this cubic and report the roots.

But solving the cubic t³ -8t -8 = 0. Let's try to find its roots.

First, check for rational roots using Rational Root Theorem. Possible rational roots are ±1, ±2, ±4, ±8. Let's test t=2:

2³ -8*2 -8 = 8 -16 -8 = -16 ≠ 0.

t=4: 64 -32 -8=24≠0.

t=1:1 -8 -8=-15≠0.

t=-1:-1 +8 -8=-1≠0.

t=-2:-8 +16 -8=0. Oh! t=-2 is a root.

Therefore, we can factor the cubic as (t + 2)(t² - 2t -4) = 0.

Indeed:

(t + 2)(t² - 2t -4) = t³ -2t² -4t + 2t² -4t -8 = t³ -8t -8. Correct.

Therefore, the roots are t = -2, and the roots of t² -2t -4 =0, which are t = [2 ± sqrt(4 +16)]/2 = [2 ± sqrt(20)]/2 = [2 ± 2 sqrt(5)]/2 = 1 ± sqrt(5).

Therefore, the resolvent cubic has roots -2, 1 + sqrt(5), and 1 - sqrt(5). Therefore, the set {S₁, S₂, S₃} is {-2, 1 + sqrt(5), 1 - sqrt(5)}.

Therefore, the answer is the set {-2, 1 - sqrt(5), 1 + sqrt(5)}.

**Final Answer**
The unordered set is \boxed{\left\{ -2, 1 - \sqrt{5}, 1 + \sqrt{5} \right\}}.
Therefore, the unordered set \(\{\alpha_1\alpha_2 + \alpha_3\alpha_4, \alpha_1\alpha_3 + \alpha_2\alpha_4, \alpha_1\alpha_4 + \alpha_2\alpha_3\}\) is equal to the set of roots of the equation \(t^3 - 8t - 8 = 0\), which are \(-2\), \(1 - \sqrt{5}\), and \(1 + \sqrt{5}\). Thus, the final answer is \(\boxed{\left\{ -2, 1 - \sqrt{5}, 1 + \sqrt{5} \right\}}\).
</think>