Explanation:
Okay, so I need to find the unordered set containing the three expressions: α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, and α₁α₄ + α₂α₃, where α₁, α₂, α₃, α₄ are the roots of the equation x⁴ + 2x³ + 2 = 0.

First, I remember that for polynomials, there are relationships between the coefficients and the roots, known as Vieta's formulas. Since this is a quartic equation, Vieta's formulas will relate the sums and products of the roots to the coefficients of the polynomial. Let me recall Vieta's formulas for a quartic equation.

For a general quartic equation x⁴ + a x³ + b x² + c x + d = 0, the sums are:

- α₁ + α₂ + α₃ + α₄ = -a
- α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = b
- α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -c
- α₁α₂α₃α₄ = d

But in our case, the equation is x⁴ + 2x³ + 2 = 0. Comparing with the general form, we have:

- a = 2, so the sum of roots α₁ + α₂ + α₃ + α₄ = -2
- b = 0 (since there is no x² term), so the sum of products two at a time is 0
- c = 0 (since there is no x term), so the sum of products three at a time is 0
- d = 2, so the product of all roots is 2

So, summarizing:

Sum of roots: S₁ = α₁ + α₂ + α₃ + α₄ = -2

Sum of products two at a time: S₂ = α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = 0

Sum of products three at a time: S₃ = α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = 0

Product of roots: S₄ = α₁α₂α₃α₄ = 2

Now, the problem asks for the set {α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃}. Let me denote these three expressions as follows:

Let’s call them:

A = α₁α₂ + α₃α₄

B = α₁α₃ + α₂α₄

C = α₁α₄ + α₂α₃

So, the set we need is {A, B, C}.

I need to find the values of A, B, C. Since the polynomial is quartic, and the roots are distinct, these expressions are symmetric in some way. Also, note that the sum of A, B, C is equal to S₂, which is 0. Because:

A + B + C = (α₁α₂ + α₃α₄) + (α₁α₃ + α₂α₄) + (α₁α₄ + α₂α₃) = (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄) = S₂ = 0.

Therefore, A + B + C = 0. So, the three numbers A, B, C add up to zero. Therefore, if I can find two of them, the third can be determined. But since the problem asks for the unordered set, maybe all three are equal? Or maybe they are different? Let me check.

Alternatively, perhaps there is a way to relate A, B, C with other symmetric sums. Let me think.

Alternatively, maybe I can consider that if I have the roots α₁, α₂, α₃, α₄, then perhaps I can pair them in different ways. However, since the polynomial is quartic, maybe there's a way to factor it or find relations between the roots?

Alternatively, perhaps I can consider that if I take two roots and multiply them, and then take the other two roots and multiply them, then add them. So, A = α₁α₂ + α₃α₄. Similarly for B and C.

But how can I find these values? Let me think. Since we know S₁, S₂, S₃, S₄, maybe we can find expressions for A, B, C. Let me think about how to relate them.

Let me note that if I consider the product of two roots, say α₁α₂, then perhaps I can relate it to other terms. But since we have symmetric sums, maybe we can find expressions for A, B, C in terms of S₁, S₂, S₃, S₄.

Alternatively, perhaps I can consider that if I take the product (α₁ + α₂)(α₃ + α₄). Let me compute that:

(α₁ + α₂)(α₃ + α₄) = α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ = B + C.

Similarly, if I take (α₁ + α₃)(α₂ + α₄) = α₁α₂ + α₁α₄ + α₃α₂ + α₃α₄ = A + C.

Similarly, (α₁ + α₄)(α₂ + α₃) = α₁α₂ + α₁α₃ + α₄α₂ + α₄α₃ = A + B.

So, if I denote:

Let me define:

Let’s say that:

Let’s denote p = α₁ + α₂, q = α₃ + α₄. Then, p + q = S₁ = -2.

Similarly, (α₁ + α₂)(α₃ + α₄) = p q = B + C.

Similarly, if I take other pairings, but maybe I need more relations.

Alternatively, perhaps I can find expressions for A, B, C. Let me think.

Alternatively, since A + B + C = 0, and if I can find AB + BC + CA, and ABC, then I can find the values of A, B, C as roots of a cubic equation. But how?

Alternatively, maybe I can find AB + BC + CA. Let me try.

First, let me compute AB + BC + CA. Let me compute AB + BC + CA.

But AB + BC + CA = A(B + C) + BC. Since A + B + C = 0, then B + C = -A. Therefore, AB + BC + CA = A(-A) + BC = -A² + BC. Similarly, but I need to find BC. Wait, but maybe this approach is getting too convoluted. Let me think again.

Alternatively, perhaps I can find expressions for A, B, C in terms of the known symmetric sums. Let me think.

Let me note that A = α₁α₂ + α₃α₄. Similarly, if I consider that the product of all roots is S₄ = 2. So, α₁α₂α₃α₄ = 2. Therefore, if I take A = α₁α₂ + α₃α₄, then if I multiply A by α₁α₂α₃α₄, but that might not help. Wait, maybe not.

Alternatively, perhaps I can consider that if I take the product of two roots, say α₁α₂, then the product of the other two roots is α₃α₄. Let me denote A = α₁α₂ + α₃α₄. Similarly, if I denote B = α₁α₃ + α₂α₄, and C = α₁α₄ + α₂α₃.

Alternatively, perhaps I can consider that if I take the square of the sum of roots:

(α₁ + α₂ + α₃ + α₄)² = α₁² + α₂² + α₃² + α₄² + 2(α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)

We know that S₁ = -2, so S₁² = 4. Also, S₂ = 0, so:

(-2)² = (sum of squares) + 2*(0) => 4 = sum of squares. Therefore, sum of squares of roots is 4.

But how does that help? Maybe not directly. Let me think.

Alternatively, perhaps I can consider the equations involving A, B, C. Let me think about the following:

Let me consider that if I take the product of two roots, say α₁α₂, then if I can relate it to other terms. For example, if I take α₁α₂, then since α₁ + α₂ + α₃ + α₄ = -2, then α₁ + α₂ = -2 - (α₃ + α₄). Let me denote p = α₁ + α₂, q = α₃ + α₄, so p + q = -2. Then, A = α₁α₂ + α₃α₄. Let me denote that as A = (α₁α₂) + (α₃α₄). Also, if I consider that α₁α₂ can be expressed in terms of p and something else. Wait, for example, if I know p = α₁ + α₂, then α₁α₂ can be found if I know something else. But how?

Alternatively, if I consider that α₁ and α₂ are roots of a quadratic equation, say x² - px + r = 0, where r = α₁α₂. Similarly, α₃ and α₄ are roots of another quadratic equation x² - qx + s = 0, where s = α₃α₄. Then, since the original quartic can be factored as (x² - px + r)(x² - qx + s) = 0. Let me try expanding this:

(x² - px + r)(x² - qx + s) = x⁴ - (p + q)x³ + (pq + r + s)x² - (ps + rq)x + rs.

Comparing with the original equation x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore, we can equate coefficients:

1. Coefficient of x⁴: 1 = 1, which is okay.

2. Coefficient of x³: -(p + q) = 2 => p + q = -2. Which matches our previous result since p + q = -2.

3. Coefficient of x²: pq + r + s = 0.

4. Coefficient of x: -(ps + rq) = 0 => ps + rq = 0.

5. Constant term: rs = 2.

So, we have these equations:

p + q = -2,

pq + r + s = 0,

ps + rq = 0,

rs = 2.

But we also know that A = r + s, since A = α₁α₂ + α₃α₄ = r + s. Similarly, if we can find r + s, which is A, and also other relations.

But we also have that p + q = -2, and pq + (r + s) = 0 => pq + A = 0 => pq = -A.

Also, from the fourth equation: ps + rq = 0. Let me note that p = α₁ + α₂, q = α₃ + α₄, r = α₁α₂, s = α₃α₄.

But how can I relate ps + rq? Let me think. Let me write ps + rq = p s + r q. Since p = α₁ + α₂, q = α₃ + α₄, r = α₁α₂, s = α₃α₄. So:

ps + rq = (α₁ + α₂)(α₃α₄) + (α₁α₂)(α₃ + α₄) = α₁α₃α₄ + α₂α₃α₄ + α₁α₂α₃ + α₁α₂α₄ = α₃α₄(α₁ + α₂) + α₁α₂(α₃ + α₄) = q s + p r. Wait, but that's the same as before. Wait, but we have ps + rq = 0. So, that's equal to q s + p r = 0.

But we also know that rs = 2. So, s = 2 / r. Therefore, substituting into the equation q s + p r = 0:

q*(2 / r) + p r = 0 => 2q / r + p r = 0 => Multiply both sides by r:

2q + p r² = 0 => p r² = -2q.

But we also have that pq = -A. So, p q = -A. Let me note that. Also, since p + q = -2, we can write q = -2 - p. Therefore, substituting into p q = -A:

p(-2 - p) = -A => -2p - p² = -A => A = 2p + p².

Similarly, from the equation p r² = -2q, and since q = -2 - p, then:

p r² = -2(-2 - p) = 4 + 2p => p r² = 4 + 2p => r² = (4 + 2p)/p = 4/p + 2.

But also, since rs = 2, and s = 2 / r, so s = 2 / r. Also, from pq + r + s = 0, which is pq + A = 0, since A = r + s. Wait, pq + A = 0 => pq = -A. But we already have that.

Alternatively, since A = r + s, and rs = 2, then r and s are roots of the quadratic equation t² - A t + 2 = 0. Therefore, r and s satisfy t² - A t + 2 = 0. Therefore, their product is 2, and their sum is A.

But also, from earlier, we have that r² = 4/p + 2. But since r is a root of t² - A t + 2 = 0, then r² = A r - 2. Therefore, equating the two expressions for r²:

A r - 2 = 4/p + 2 => A r = 4/p + 4 => A r = 4(1/p + 1) => A r = 4( (1 + p)/p )

But since p + q = -2, and q = -2 - p, and we also have that pq = -A. Let me recall that pq = -A. So, p(-2 - p) = -A => -2p - p² = -A => A = 2p + p². Therefore, A = p² + 2p.

So, substituting A into the previous equation:

A r = 4( (1 + p)/p )

=> (p² + 2p) r = 4( (1 + p)/p )

But since r is a root of t² - A t + 2 = 0, which is t² - (p² + 2p) t + 2 = 0. Therefore, r² = (p² + 2p) r - 2. But we also have that r² = 4/p + 2. Therefore, equating:

(p² + 2p) r - 2 = 4/p + 2 => (p² + 2p) r = 4/p + 4 => same as before.

But maybe I can express r in terms of p? Let me think. Since r is a root of t² - A t + 2 = 0, and A = p² + 2p, then:

t² - (p² + 2p) t + 2 = 0.

But also, from rs = 2, so s = 2 / r. Also, since A = r + s = r + 2 / r. Therefore, A = r + 2 / r. Therefore, multiplying both sides by r:

A r = r² + 2. But from the quadratic equation, r² = A r - 2. Therefore, substituting into A r = r² + 2 gives A r = (A r - 2) + 2 => A r = A r. Which is an identity. So, not helpful.

Hmm. Maybe I need another approach. Let me think again.

We have that A = p² + 2p, where p = α₁ + α₂. Also, since p + q = -2, q = -2 - p. Also, pq = -A. Therefore, p(-2 - p) = -A => A = 2p + p².

Additionally, we have that from the equation p r² = 4 + 2p. But since r² = A r - 2, substituting:

p(A r - 2) = 4 + 2p => p A r - 2p = 4 + 2p => p A r = 4 + 4p => A r = (4 + 4p)/p = 4(1 + p)/p.

But since A = p² + 2p, then:

(p² + 2p) r = 4(1 + p)/p => r = [4(1 + p)/p] / (p² + 2p) = [4(1 + p)/p] / [p(p + 2)] = 4(1 + p) / [p²(p + 2)].

But also, since A = r + 2 / r, we can write:

A = r + 2 / r. Let me substitute r from above:

A = [4(1 + p)/ (p²(p + 2))] + 2 / [4(1 + p)/ (p²(p + 2))]

Simplify the second term:

2 divided by [4(1 + p)/ (p²(p + 2))] = 2 * [p²(p + 2) / (4(1 + p))] = [p²(p + 2)] / [2(1 + p)]

Therefore, A = [4(1 + p)/ (p²(p + 2))] + [p²(p + 2)] / [2(1 + p)]

This seems complicated, but maybe we can combine these terms. Let me denote:

Let me write A as:

A = [4(1 + p)] / [p²(p + 2)] + [p²(p + 2)] / [2(1 + p)]

Let me find a common denominator. The denominators are p²(p + 2) and 2(1 + p). The common denominator would be 2p²(p + 2)(1 + p).

So, first term becomes:

[4(1 + p)] / [p²(p + 2)] * [2(1 + p)] / [2(1 + p)] = [8(1 + p)^2] / [2p²(p + 2)(1 + p)]

Wait, no. Wait, to get the common denominator, multiply numerator and denominator by 2(1 + p):

First term: [4(1 + p)] / [p²(p + 2)] * [2(1 + p)] / [2(1 + p)] = [4(1 + p) * 2(1 + p)] / [2p²(p + 2)(1 + p)] = [8(1 + p)^2] / [2p²(p + 2)(1 + p)] = [4(1 + p)] / [p²(p + 2)]

Wait, that seems like going in circles. Maybe another approach. Let me denote t = p. Then, A = t² + 2t. Also, from above, A = [4(1 + t)/ (t²(t + 2))] + [t²(t + 2)] / [2(1 + t)]. Let me compute this expression:

Let me compute the two terms:

Term1 = 4(1 + t)/[t²(t + 2)]

Term2 = [t²(t + 2)]/[2(1 + t)]

So, A = Term1 + Term2. But since A = t² + 2t, we can set up the equation:

t² + 2t = Term1 + Term2.

This seems complicated, but maybe we can substitute t² + 2t = A and see if we can find t. However, this seems too involved. Maybe there's a better way.

Alternatively, since we have that rs = 2 and A = r + s, then r and s are roots of t² - A t + 2 = 0. Also, from earlier, we have that p r² = 4 + 2p. But since r² = A r - 2, then:

p(A r - 2) = 4 + 2p => p A r - 2p = 4 + 2p => p A r = 4 + 4p => A r = (4 + 4p)/p = 4(1 + p)/p.

But since A = t² + 2t where t = p, and A = r + s, and r is a root of t² - A t + 2 = 0, then r can be expressed as [A ± sqrt(A² - 8)] / 2. But this seems too abstract.

Alternatively, maybe I can consider that since A = r + s and rs = 2, then A is the sum of two numbers whose product is 2. Therefore, A can be any real or complex number, but given that the original equation has complex roots, A could be complex. However, since the problem asks for the unordered set {A, B, C}, which are symmetric expressions, maybe they are all equal? Or maybe they are roots of some equation?

Alternatively, since A + B + C = 0, and if I can find AB + BC + CA and ABC, then I can find the cubic equation whose roots are A, B, C. Then, solving that equation would give me the values of A, B, C.

So, let me try to compute AB + BC + CA and ABC.

First, AB + BC + CA. Let me recall that:

AB + BC + CA = (A + B + C)^2 - (A² + B² + C²) all over 2? Wait, no. Wait, (A + B + C)^2 = A² + B² + C² + 2(AB + BC + CA). Therefore, AB + BC + CA = [(A + B + C)^2 - (A² + B² + C²)] / 2. But since A + B + C = 0, this simplifies to AB + BC + CA = - (A² + B² + C²)/2. But I don't know A² + B² + C².

Alternatively, maybe I can find AB + BC + CA directly. Let me think.

Let me note that A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me compute AB.

AB = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Let me expand this:

= α₁α₂α₁α₃ + α₁α₂α₂α₄ + α₃α₄α₁α₃ + α₃α₄α₂α₄

= α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²

Hmm, this seems complicated. Maybe there's a better way. Alternatively, perhaps I can use symmetric sums.

Alternatively, note that AB + BC + CA can be expressed in terms of the symmetric sums. Let me think.

Alternatively, consider that AB + BC + CA = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄) + (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃) + (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄).

This seems very complex, but maybe there's a pattern. Alternatively, perhaps I can use the fact that A + B + C = 0 and find AB + BC + CA by considering other relations.

Alternatively, let me think about the product of the roots. Since α₁α₂α₃α₄ = 2. Let me note that:

Let me consider that A = α₁α₂ + α₃α₄. Let me denote that as A. Similarly, if I multiply A by α₁α₂α₃α₄, which is 2, then:

A * 2 = (α₁α₂ + α₃α₄) * 2 = 2α₁α₂ + 2α₃α₄.

But not sure if helpful.

Alternatively, let me think about the following. Let me consider that if I take the product of all three expressions A, B, C. But that might be too complicated.

Alternatively, maybe I can use Newton's identities? But I'm not sure.

Alternatively, maybe I can consider that since the original equation is x⁴ + 2x³ + 2 = 0, maybe I can find relations between the roots. Let me try to factor the polynomial.

Let me try to factor x⁴ + 2x³ + 2. Maybe it factors into quadratics? Let me assume that it factors as (x² + a x + b)(x² + c x + d). Then expanding:

x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + b d.

Comparing with original equation x⁴ + 2x³ + 0x² + 0x + 2. Therefore, we have the following equations:

1. a + c = 2

2. ac + b + d = 0

3. ad + bc = 0

4. b d = 2

So, similar to earlier equations. Let me try to solve these equations.

From equation 1: c = 2 - a.

From equation 4: b d = 2. Let me assume that b and d are real numbers? Not necessarily, since the roots are complex. But maybe we can find integer solutions? Let me check.

Possible integer pairs for b and d such that b d = 2: (1, 2), (2, 1), (-1, -2), (-2, -1). Let me try these possibilities.

First, let me try b = 1, d = 2. Then equation 3: a*2 + b*c = 0 => 2a + 1*c = 0. But c = 2 - a, so 2a + (2 - a) = 0 => a + 2 = 0 => a = -2. Then c = 2 - (-2) = 4. Then check equation 2: ac + b + d = (-2)(4) + 1 + 2 = -8 + 3 = -5 ≠ 0. Not good.

Next, try b = 2, d = 1. Then equation 3: a*1 + 2*c = 0 => a + 2c = 0. Since c = 2 - a, substitute: a + 2(2 - a) = 0 => a + 4 - 2a = 0 => -a + 4 = 0 => a = 4. Then c = 2 - 4 = -2. Check equation 2: ac + b + d = (4)(-2) + 2 + 1 = -8 + 3 = -5 ≠ 0. Not good.

Next, try b = -1, d = -2. Equation 3: a*(-2) + (-1)*c = 0 => -2a - c = 0 => c = -2a. But c = 2 - a, so -2a = 2 - a => -2a + a = 2 => -a = 2 => a = -2. Then c = 2 - (-2) = 4. Check equation 2: ac + b + d = (-2)(4) + (-1) + (-2) = -8 - 3 = -11 ≠ 0. Not good.

Next, try b = -2, d = -1. Equation 3: a*(-1) + (-2)*c = 0 => -a - 2c = 0 => a = -2c. But c = 2 - a, so substitute: a = -2(2 - a) => a = -4 + 2a => -a = -4 => a = 4. Then c = 2 - 4 = -2. Check equation 2: ac + b + d = (4)(-2) + (-2) + (-1) = -8 - 3 = -11 ≠ 0. Not good.

So none of the integer pairs work. Maybe the factors are not with integer coefficients. Let me try to think differently.

Alternatively, maybe the quartic can be written as x⁴ + 2x³ + 2 = 0. Let me try to make a substitution. Let me set y = x + k, to eliminate the cubic term? Wait, but the cubic term is already present. Alternatively, maybe divide by x²? Let me try:

x⁴ + 2x³ + 2 = 0 => x² + 2x + 2/x² = 0. Let me set z = x + 1/x. Then z² = x² + 2 + 1/x². Therefore, x² + 1/x² = z² - 2. But our equation is x² + 2x + 2/x² = 0. Let me write it as (x² + 2/x²) + 2x = 0. But x² + 2/x² = (x² + 1/x²) + 1/x² = (z² - 2) + 1/x². Not sure. Alternatively, maybe not helpful.

Alternatively, maybe write the equation as x⁴ + 2x³ = -2. Then factor x³(x + 2) = -2. Not sure.

Alternatively, maybe consider that the equation is x⁴ + 2x³ + 2 = 0. Let me try to find roots numerically? But since the problem is about symmetric expressions, maybe not helpful.

Alternatively, maybe consider that since the polynomial is x⁴ + 2x³ + 2, and we know Vieta's formulas, maybe we can find relations between A, B, C.

Earlier, I had that A + B + C = 0. Also, if I can find AB + BC + CA and ABC, then I can find the cubic equation whose roots are A, B, C.

Let me try to compute AB + BC + CA.

Let me recall that A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me compute AB + BC + CA.

First, compute AB:

AB = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄)

= α₁α₂α₁α₃ + α₁α₂α₂α₄ + α₃α₄α₁α₃ + α₃α₄α₂α₄

= α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²

Similarly, compute BC:

BC = (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃)

= α₁α₃α₁α₄ + α₁α₃α₂α₃ + α₂α₄α₁α₄ + α₂α₄α₂α₃

= α₁²α₃α₄ + α₁α₂α₃² + α₁α₂α₄² + α₂²α₃α₄

Similarly, compute CA:

CA = (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄)

= α₁α₄α₁α₂ + α₁α₄α₃α₄ + α₂α₃α₁α₂ + α₂α₃α₃α₄

= α₁²α₂α₄ + α₁α₃α₄² + α₁α₂²α₃ + α₂α₃²α₄

Now, adding AB + BC + CA:

Let me list all terms from AB, BC, CA:

From AB:

1. α₁²α₂α₃

2. α₁α₂²α₄

3. α₁α₃²α₄

4. α₂α₃α₄²

From BC:

5. α₁²α₃α₄

6. α₁α₂α₃²

7. α₁α₂α₄²

8. α₂²α₃α₄

From CA:

9. α₁²α₂α₄

10. α₁α₃α₄²

11. α₁α₂²α₃

12. α₂α₃²α₄

Now, let me group similar terms:

Terms with α₁²α₂α₃: terms 1 and 11. Term 1 is α₁²α₂α₃, term 11 is α₁α₂²α₃. Wait, no, term 11 is α₁α₂²α₃? Wait, term 11 is from CA: α₁α₂²α₃? Let me check:

From CA: term 11 is α₁α₂²α₃? Wait, no. Let me recheck:

From CA:

Term 9: α₁²α₂α₄

Term 10: α₁α₃α₄²

Term 11: α₁α₂²α₃

Term 12: α₂α₃²α₄

So, term 11 is α₁α₂²α₃, term 1 is α₁²α₂α₃. So, terms 1 and 11 are α₁²α₂α₃ and α₁α₂²α₃. Similarly, terms 2 and 6: term 2 is α₁α₂²α₄, term 6 is α₁α₂α₃². Wait, no. Let me check:

Wait, term 2 is α₁α₂²α₄, term 6 is α₁α₂α₃². Not the same.

This seems too messy. Maybe there's a better way. Let me think about the sum AB + BC + CA.

Alternatively, note that AB + BC + CA can be expressed as:

AB + BC + CA = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄) + (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃) + (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄)

Let me denote that each term is a product of two of the expressions A, B, C. But since A + B + C = 0, maybe there's a relation.

Alternatively, consider that AB + BC + CA = (A + B + C)^2 - (A² + B² + C²) all over 2, but since A + B + C = 0, this becomes - (A² + B² + C²)/2. But I don't know A² + B² + C².

Alternatively, maybe I can find A² + B² + C². Let me try:

A² + B² + C² = (A + B + C)^2 - 2(AB + BC + CA) = 0 - 2(AB + BC + CA) = -2(AB + BC + CA).

But this seems circular.

Alternatively, maybe I can find A² + B² + C² by expanding A², B², C².

Let me compute A² = (α₁α₂ + α₃α₄)^2 = α₁²α₂² + 2α₁α₂α₃α₄ + α₃²α₄². Similarly for B² and C².

So, A² + B² + C² = (α₁²α₂² + α₃²α₄²) + (α₁²α₃² + α₂²α₄²) + (α₁²α₄² + α₂²α₃²) + 6α₁α₂α₃α₄.

Wait, because each A², B², C² has a term 2α₁α₂α₃α₄, so total of 3*2α₁α₂α₃α₄ = 6α₁α₂α₃α₄.

So, A² + B² + C² = [α₁²α₂² + α₃²α₄² + α₁²α₃² + α₂²α₄² + α₁²α₄² + α₂²α₃²] + 6α₁α₂α₃α₄.

Now, note that α₁²α₂² + α₁²α₃² + α₁²α₄² + α₂²α₃² + α₂²α₄² + α₃²α₄² = (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 - 2(α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄)(α₁ + α₂ + α₃ + α₄) + 2α₁α₂α₃α₄. Wait, this seems too complicated.

Alternatively, recall that (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = S₂² = 0² = 0.

But expanding this square gives:

(α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = Σα_i²α_j² + 2Σα_iα_jα_kα_l where i < j, k < l. Wait, no. Wait, more carefully:

When you square the sum of products two at a time, you get the sum of squares of each product plus twice the sum of products of distinct pairs of products.

So, Σ(α_iα_j)^2 + 2Σ_{i < j, k < l, (i,j) ≠ (k,l)} α_iα_jα_kα_l.

But this seems complex. However, we know that Σ(α_iα_j)^2 = A² + B² + C² - 6α₁α₂α₃α₄? Wait, no. Earlier, we had:

A² + B² + C² = Σ(α_iα_j)^2 + 6α₁α₂α₃α₄.

Wait, earlier, I had:

A² + B² + C² = [Σ(α_iα_j)^2] + 6α₁α₂α₃α₄.

But since S₂ = Σ(α_iα_j) = 0, then S₂² = [Σ(α_iα_j)]² = 0 = Σ(α_iα_j)^2 + 2Σ_{i < j, k < l} α_iα_jα_kα_l.

But Σ_{i < j, k < l} α_iα_jα_kα_l is equal to the sum over all products of two distinct pairs. However, this is equal to (Σα_iα_j)(Σα_kα_l) - Σ(α_iα_j)^2. Wait, no. Wait, if I consider (Σα_iα_j)(Σα_kα_l) = Σα_iα_jα_kα_l + Σα_iα_jα_kα_l where the pairs overlap. Wait, this is getting too tangled.

Alternatively, since we know that Σ(α_iα_j)^2 = S₂² - 2Σα_iα_jα_kα_l, but I'm not sure.

Alternatively, since we know that α₁α₂α₃α₄ = 2, so the product of all roots is 2. Let me think about Σ(α_iα_j)^2.

Let me note that Σ(α_iα_j)^2 = (α₁α₂)^2 + (α₁α₃)^2 + (α₁α₄)^2 + (α₂α₃)^2 + (α₂α₄)^2 + (α₃α₄)^2.

Let me consider that (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = 0 = Σ(α_iα_j)^2 + 2Σ_{i < j, k < l} α_iα_jα_kα_l.

But Σ_{i < j, k < l} α_iα_jα_kα_l is equal to the sum over all products of two distinct pairs. How many terms are there? For four variables, the number of ways to choose two distinct pairs is C(6, 2) minus the cases where the pairs overlap. Wait, no. Actually, the number of ways to choose two distinct pairs is C(6, 2), but many of these pairs overlap.

Alternatively, note that Σ_{i < j, k < l} α_iα_jα_kα_l = (Σα_iα_j)(Σα_kα_l) - Σ(α_iα_j)^2. Wait, no. Wait, if I multiply Σα_iα_j by Σα_kα_l, I get Σα_iα_jα_kα_l + Σα_iα_jα_kα_l where the pairs (i,j) and (k,l) overlap. But this is too vague.

Alternatively, since we know that Σα_iα_j = 0, and Σα_iα_jα_k = 0, and Σα_iα_jα_kα_l = 2.

Wait, let me think. The product of all four roots is 2. So, if I take the product of two pairs, say α₁α₂α₃α₄ = 2. So, for any two pairs, their product is 2. For example, (α₁α₂)(α₃α₄) = 2, (α₁α₃)(α₂α₄) = 2, (α₁α₄)(α₂α₃) = 2. Therefore, each of the products of two pairs is 2.

Therefore, Σ_{i < j, k < l} α_iα_jα_kα_l = 3 * 2 = 6? Wait, no. Wait, there are three ways to split the four roots into two pairs: (α₁α₂)(α₃α₄), (α₁α₃)(α₂α₄), (α₁α₄)(α₂α₃). Each of these products is equal to 2. Therefore, the sum of all such products is 3 * 2 = 6.

But earlier, we had that (Σα_iα_j)^2 = Σ(α_iα_j)^2 + 2 * Σ_{i < j, k < l} α_iα_jα_kα_l.

Since Σα_iα_j = 0, then 0 = Σ(α_iα_j)^2 + 2 * 6 => Σ(α_iα_j)^2 = -12.

Wait, this is interesting! So, Σ(α_iα_j)^2 = -12.

Therefore, going back to A² + B² + C² = Σ(α_iα_j)^2 + 6α₁α₂α₃α₄ = -12 + 6*2 = -12 + 12 = 0.

So, A² + B² + C² = 0.

But earlier, we had that AB + BC + CA = - (A² + B² + C²)/2 = -0/2 = 0.

So, AB + BC + CA = 0.

Therefore, now we know that A + B + C = 0, AB + BC + CA = 0, and we need to find ABC.

If we can find ABC, then we can write the cubic equation whose roots are A, B, C as x³ - (A + B + C)x² + (AB + BC + CA)x - ABC = x³ - 0x² + 0x - ABC = x³ - ABC = 0. Therefore, the cubic equation is x³ - ABC = 0, so the roots are the cube roots of ABC.

But since A, B, C are complex numbers, and the original equation has complex roots, it's possible that ABC is a real number or complex. Let me try to find ABC.

Let me compute ABC.

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

This seems complicated, but maybe there's a way to simplify it. Let me think.

Let me recall that earlier, we had that each pair product is 2. For example, (α₁α₂)(α₃α₄) = 2, (α₁α₃)(α₂α₄) = 2, (α₁α₄)(α₂α₃) = 2.

Let me denote that:

Let me denote A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me consider multiplying A, B, C.

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

Let me first compute (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Earlier, we expanded this and got terms like α₁²α₂α₃ + ... but maybe there's a smarter way.

Alternatively, let me note that if I let x = α₁α₂, y = α₃α₄, z = α₁α₃, w = α₂α₄, v = α₁α₄, u = α₂α₃. Then, A = x + y, B = z + w, C = v + u. But this might not help.

Alternatively, let me consider that since (α₁α₂)(α₃α₄) = 2, so x y = 2, similarly z w = 2, v u = 2.

But how does this help?

Alternatively, let me consider that ABC = (A)(B)(C). Let me try to compute this product.

Let me first compute (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Let me denote this as (x + y)(z + w), where x = α₁α₂, y = α₃α₄, z = α₁α₃, w = α₂α₄.

Expanding this gives xz + xw + yz + yw.

Now, xz = α₁α₂ * α₁α₃ = α₁²α₂α₃.

xw = α₁α₂ * α₂α₄ = α₁α₂²α₄.

yz = α₃α₄ * α₁α₃ = α₁α₃²α₄.

yw = α₃α₄ * α₂α₄ = α₂α₃α₄².

So, the product (x + y)(z + w) = α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄².

Now, multiply this by (v + u) = α₁α₄ + α₂α₃. Let me denote this as (v + u).

So, ABC = [α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²] * (α₁α₄ + α₂α₃).

This seems very complex, but maybe we can find a pattern or use known values.

Let me note that α₁α₂α₃α₄ = 2. So, for example, α₁α₂α₃ = 2 / α₄, and similarly for other terms. Let me try to express each term in the first bracket in terms of α₁α₂α₃α₄.

For example, α₁²α₂α₃ = α₁ * (α₁α₂α₃) = α₁ * (2 / α₄). Similarly, α₁α₂²α₄ = α₂ * (α₁α₂α₄) = α₂ * (2 / α₃). Similarly, α₁α₃²α₄ = α₃ * (α₁α₃α₄) = α₃ * (2 / α₂). And α₂α₃α₄² = α₄ * (α₂α₃α₄) = α₄ * (2 / α₁).

Therefore, the first bracket becomes:

α₁*(2 / α₄) + α₂*(2 / α₃) + α₃*(2 / α₂) + α₄*(2 / α₁) = 2[α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁].

Now, multiplying this by (α₁α₄ + α₂α₃):

ABC = 2[α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁] * (α₁α₄ + α₂α₃).

Let me compute this product. Let me denote the first bracket as 2K, where K = α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁.

So, ABC = 2K * (α₁α₄ + α₂α₃).

Let me compute K * (α₁α₄ + α₂α₃).

Let me expand this:

K * (α₁α₄ + α₂α₃) = [α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁] * (α₁α₄ + α₂α₃).

Let me distribute each term:

First term: α₁/α₄ * α₁α₄ = α₁²

Second term: α₁/α₄ * α₂α₃ = α₁α₂α₃ / α₄

Third term: α₂/α₃ * α₁α₄ = α₁α₂α₄ / α₃

Fourth term: α₂/α₃ * α₂α₃ = α₂²

Fifth term: α₃/α₂ * α₁α₄ = α₁α₃α₄ / α₂

Sixth term: α₃/α₂ * α₂α₃ = α₃²

Seventh term: α₄/α₁ * α₁α₄ = α₄²

Eighth term: α₄/α₁ * α₂α₃ = α₂α₃α₄ / α₁

So, combining all terms:

α₁² + α₂² + α₃² + α₄² + [α₁α₂α₃ / α₄ + α₁α₂α₄ / α₃ + α₁α₃α₄ / α₂ + α₂α₃α₄ / α₁]

Now, let me simplify each of the bracketed terms.

Note that α₁α₂α₃ / α₄ = (α₁α₂α₃α₄) / α₄² = 2 / α₄². Similarly:

α₁α₂α₃ / α₄ = 2 / α₄²

α₁α₂α₄ / α₃ = 2 / α₃²

α₁α₃α₄ / α₂ = 2 / α₂²

α₂α₃α₄ / α₁ = 2 / α₁²

Therefore, the bracketed terms become 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

So, overall, K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Therefore, ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²) ].

Now, we need to compute α₁² + α₂² + α₃² + α₄² and 1/α₁² + 1/α₂² + 1/α₃² + 1/α₄².

Earlier, we found that the sum of squares of the roots is 4. Let me recall that:

From earlier, we had that (α₁ + α₂ + α₃ + α₄)^2 = 4 = sum of squares + 2 * sum of products two at a time. Since sum of products two at a time is 0, then sum of squares is 4. So, α₁² + α₂² + α₃² + α₄² = 4.

Now, let me compute 1/α₁² + 1/α₂² + 1/α₃² + 1/α₄².

Let me denote this as S = Σ1/α_i².

Note that Σ1/α_i² = (Σ1/α_i)^2 - 2Σ1/(α_iα_j).

Let me compute Σ1/α_i and Σ1/(α_iα_j).

First, Σ1/α_i = (α₂α₃α₄ + α₁α₃α₄ + α₁α₂α₄ + α₁α₂α₃) / (α₁α₂α₃α₄).

From Vieta's formulas, the sum of products three at a time is S₃ = 0, and the product of all roots is S₄ = 2. Therefore, Σ1/α_i = S₃ / S₄ = 0 / 2 = 0.

Next, Σ1/(α_iα_j) = Σ(α_kα_l) / (α_iα_jα_kα_l) where α_kα_l is the product of the other two roots. Wait, let me think.

Σ1/(α_iα_j) = Σ_{i < j} 1/(α_iα_j).

But since α₁α₂α₃α₄ = 2, then 1/(α_iα_j) = α_kα_l / 2, where α_kα_l is the product of the other two roots.

Therefore, Σ1/(α_iα_j) = (α₃α₄ + α₁α₄ + α₁α₃ + α₂α₃ + α₂α₄ + α₁α₂)/2 = S₂ / 2 = 0 / 2 = 0.

Therefore, Σ1/α_i² = (Σ1/α_i)^2 - 2Σ1/(α_iα_j) = 0² - 2*0 = 0.

Wait, that's interesting! So, Σ1/α_i² = 0.

Therefore, going back to ABC:

ABC = 2 * [4 + 2*0] = 2 * 4 = 8.

Wait, so ABC = 8?

Wait, let me check again:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ] = 2 * [4 + 2*0] = 2 * 4 = 8.

Yes, that seems correct.

Therefore, ABC = 8.

So, now we have that the cubic equation with roots A, B, C is x³ - ABC = x³ - 8 = 0. Therefore, the roots are the cube roots of 8.

But 8 is 2³, so the cube roots of 8 are 2, 2ω, 2ω², where ω is a primitive third root of unity.

But wait, the cubic equation is x³ - 8 = 0, which factors as (x - 2)(x² + 2x + 4) = 0. Therefore, the roots are 2, -1 + i√3, -1 - i√3.

But the problem says that the roots α₁, α₂, α₃, α₄ are distinct. So, the set {A, B, C} is the set containing 2, -1 + i√3, -1 - i√3.

But the problem asks for the unordered set {A, B, C}. So, the answer is {2, -1 + i√3, -1 - i√3}.

But let me check if this makes sense.

Wait, but earlier we had that A + B + C = 0, and the roots of x³ - 8 = 0 are 2, -1 ± i√3. Let me check if their sum is 0: 2 + (-1 + i√3) + (-1 - i√3) = 2 - 1 - 1 + i√3 - i√3 = 0. Yes, that works.

Also, AB + BC + CA = 0, as we found earlier. Let me check with these roots. Let me compute AB + BC + CA for roots 2, -1 + i√3, -1 - i√3.

Let me denote the roots as 2, ω, ω², where ω = -1 + i√3 and ω² = -1 - i√3.

Then AB + BC + CA = ω² + ω + 2ω² + 2ω + ω * 2? Wait, no. Wait, AB + BC + CA for the roots 2, ω, ω² is:

Let me compute AB + BC + CA where A = 2, B = ω, C = ω².

AB + BC + CA = 2ω + ω * ω² + ω² * 2 = 2ω + ω³ + 2ω².

But ω³ = 1, since ω is a primitive third root of unity. Therefore, AB + BC + CA = 2ω + 1 + 2ω².

But since ω + ω² = -1, then 2ω + 2ω² = 2(ω + ω²) = -2. Therefore, AB + BC + CA = -2 + 1 = -1. Wait, but earlier we had AB + BC + CA = 0. This is a contradiction!

Wait, this suggests an error in my calculation.

Wait, earlier I said that AB + BC + CA = 0, but according to this, it's -1. Where did I go wrong?

Let me retrace. Earlier, I had that A² + B² + C² = 0, and since A + B + C = 0, then AB + BC + CA = - (A² + B² + C²)/2 = 0. But according to the roots of x³ - 8 = 0, AB + BC + CA is not zero. So, there must be a mistake in my calculation of ABC.

Let me check the calculation of ABC again.

Earlier, I had ABC = 8, but if the roots are 2, ω, ω², then ABC = 2 * ω * ω² = 2 * ω³ = 2 * 1 = 2. But earlier I said ABC = 8. So, there's a contradiction here.

This suggests that my calculation of ABC was wrong. Let me go back.

Let me re-examine the calculation of ABC.

Earlier, I had:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ] = 2 * [4 + 2*0] = 8.

But if the roots are 2, ω, ω², then ABC should be 2 * ω * ω² = 2 * 1 = 2. So, there's a discrepancy.

Where is the mistake? Let me check the steps again.

Starting from ABC = 2 * [ (sum of squares) + 2(sum of reciprocals squared) ] = 2 * [4 + 0] = 8.

But if ABC is 2, then this suggests that my calculation is wrong.

Let me go back to the step where I calculated ABC.

I had:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ]

But let me check that step again.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Then ABC = 2 * [ ... ]

But let me check the expansion again.

When I expanded K * (α₁α₄ + α₂α₃), I got:

α₁² + α₂² + α₃² + α₄² + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But let me verify this.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But let me check with specific values. Suppose all α_i are real numbers, but in our case, they are complex. However, let me take a simple case where α₁ = α₂ = α₃ = α₄ = 1. Then, the original equation would be x⁴ + 2x³ + 2 = 0, which is not satisfied, but just for testing.

But let me not get bogged down. Let me think about the derivation again.

When I expanded K * (α₁α₄ + α₂α₃), I had:

α₁² + α₂² + α₃² + α₄² + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But how did I get that? Let me re-examine the expansion.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = α₁² + α₂² + α₃² + α₄² + [α₁α₂α₃ / α₄ + α₁α₂α₄ / α₃ + α₁α₃α₄ / α₂ + α₂α₃α₄ / α₁].

Then, I said that each of those terms is 2 / α_i².

Wait, let me check that.

For example, α₁α₂α₃ / α₄ = (α₁α₂α₃α₄) / α₄² = 2 / α₄². Yes, that's correct. Similarly for the others.

Therefore, the bracketed terms are 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Therefore, K * (α₁α₄ + α₂α₃) = (sum of squares) + 2(sum of reciprocals squared).

Then, ABC = 2 * [sum of squares + 2(sum of reciprocals squared)] = 2 * [4 + 2*0] = 8.

But if ABC is 8, then the cubic equation is x³ - 8 = 0, which has roots 2, -1 ± i√3. But then AB + BC + CA should be 0, but according to the roots, it's -1.

Wait, let me calculate AB + BC + CA for the roots 2, -1 + i√3, -1 - i√3.

Let me denote A = 2, B = -1 + i√3, C = -1 - i√3.

Then AB + BC + CA = (2)(-1 + i√3) + (-1 + i√3)(-1 - i√3) + (-1 - i√3)(2).

Compute each term:

First term: 2*(-1 + i√3) = -2 + 2i√3

Second term: (-1 + i√3)(-1 - i√3) = (-1)(-1) + (-1)(-i√3) + (i√3)(-1) + (i√3)(-i√3) = 1 + i√3 - i√3 - i²*3 = 1 - (-1)*3 = 1 + 3 = 4

Third term: (-1 - i√3)*2 = -2 - 2i√3

Adding all terms: (-2 + 2i√3) + 4 + (-2 - 2i√3) = (-2 - 2 + 4) + (2i√3 - 2i√3) = 0 + 0 = 0.

Ah! So AB + BC + CA = 0, which matches our previous result. Earlier, I must have made a mistake in calculating AB + BC + CA. So, the roots are correct.

Therefore, the cubic equation is x³ - 8 = 0, with roots 2, -1 + i√3, -1 - i√3.

Therefore, the set {A, B, C} is {2, -1 + i√3, -1 - i√3}.

But the problem says "the unordered set", so we need to present it in boxed notation. However, the problem might expect the answer in a specific form. Let me check if there's a simpler way or if I made any miscalculations.

Alternatively, maybe there's a mistake in assuming that ABC = 8. Let me verify with another approach.

Let me recall that A, B, C are the three expressions, and we have A + B + C = 0, AB + BC + CA = 0, and ABC = 8. Therefore, the cubic equation is x³ - 8 = 0, which has roots as above.

Alternatively, since the original equation is x⁴ + 2x³ + 2 = 0, and we have the roots α₁, α₂, α₃, α₄, and we are to find the set {A, B, C}.

But let me think if there's another way to find A, B, C.

Alternatively, since we know that A = α₁α₂ + α₃α₄, and we know that α₁α₂α₃α₄ = 2, so α₁α₂ = 2 / (α₃α₄). Therefore, A = 2 / (α₃α₄) + α₃α₄. Let me denote y = α₃α₄, then A = y + 2/y. Similarly, since α₁α₂ = 2/y.

Similarly, if I consider that the product of α₁α₂ and α₃α₄ is 2, so they are reciprocal up to a constant.

But since A = y + 2/y, then A is a function of y. Similarly, if I consider other pairings, B and C would be similar expressions.

But since the roots are symmetric, the values of A, B, C should be the same as the roots of the equation t = y + 2/y, where y is a product of two roots.

But since the products of two roots are α₁α₂, α₁α₃, α₁α₄, α₂α₃, α₂α₄, α₃α₄. But we know that each of these products is either y or 2/y, depending on the pairing.

But since there are three distinct pairings (as A, B, C), each corresponding to different pairings, and each of these pairings would give a value of the form y + 2/y, where y is one of the products.

But since the products come in reciprocal pairs, each value of y + 2/y would be the same for reciprocal pairs.

But since there are three distinct values of A, B, C, each corresponding to different pairings, and each of these values satisfies t = y + 2/y, where y is a product of two roots.

But since the products of two roots are related to the roots of the equation.

Alternatively, since we have already derived that the set {A, B, C} is {2, -1 + i√3, -1 - i√3}, which are the roots of x³ - 8 = 0, then this must be the correct answer.

Therefore, the unordered set is {2, -1 + i√3, -1 - i√3}.

But let me check if this makes sense with the original equation.

Let me consider that if A = 2, then α₁α₂ + α₃α₄ = 2. Since α₁α₂α₃α₄ = 2, then α₁α₂ = 2 / α₃α₄. So, if α₁α₂ + α₃α₄ = 2, then 2 / α₃α₄ + α₃α₄ = 2. Let me set z = α₃α₄, then 2/z + z = 2 => z² - 2z + 2 = 0 => z = [2 ± √(4 - 8)] / 2 = [2 ± √(-4)] / 2 = 1 ± i.

So, α₃α₄ = 1 + i or 1 - i. Similarly, the other pairings would give similar results.

But since the original equation has complex roots, this seems plausible.

Therefore, the set {A, B, C} is {2, -1 + i√3, -1 - i√3}.

But let me check if these values are consistent with the original equation.

Alternatively, since the problem asks for the unordered set, and the answer is three complex numbers, which are 2, and the two complex conjugates.

Therefore, the final answer is the set containing 2, -1 + i√3, and -1 - i√3.

But let me check if there's a simpler way to write this. For example, the problem might expect the answer in terms of radicals or something else, but since it's an unordered set, and the answer is three elements, I think this is correct.

Therefore, the answer is \boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}.

**Final Answer**
\boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}

To solve the problem, we are given the quartic polynomial:

$$
x^4 + 2x^3 + 2 = 0
$$

Let the roots of this equation be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, which are distinct complex numbers. We are asked to determine the unordered set:

$$
\{\alpha_1\alpha_2 + \alpha_3\alpha_4, \alpha_1\alpha_3 + \alpha_2\alpha_4, \alpha_1\alpha_4 + \alpha_2\alpha_3\}
$$

---

### Step 1: Use Vieta's Formulas

From the polynomial $x^4 + 2x^3 + 2 = 0$, we can apply Vieta's formulas:

- Sum of roots: $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = -2$
- Sum of products of roots taken two at a time: $\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = 0$
- Sum of products of roots taken three at a time: $\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \alpha_1\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4 = 0$
- Product of all roots: $\alpha_1\alpha_2\alpha_3\alpha_4 = 2$

Let us denote the three expressions we are interested in as:

- $A = \alpha_1\alpha_2 + \alpha_3\alpha_4$
- $B = \alpha_1\alpha_3 + \alpha_2\alpha_4$
- $C = \alpha_1\alpha_4 + \alpha_2\alpha_3$

---

### Step 2: Use Symmetry and Vieta's Identities

We observe that:

$$
A + B + C = (\alpha_1\alpha_2 + \alpha_3\alpha_4) + (\alpha_1\alpha_3 + \alpha_2\alpha_4) + (\alpha_1\alpha_4 + \alpha_2\alpha_3) = \text{sum of all products of roots two at a time} = 0
$$

So:

$$
A + B + C = 0
$$

Next, we compute $AB + BC + CA$. This is a symmetric expression that can be derived using the identities of symmetric polynomials. After a detailed expansion and simplification (as shown in the thought process), we find:

$$
AB + BC + CA = 0
$$

Finally, we compute $ABC$. Through a careful expansion and substitution using Vieta's formulas, we find:

$$
ABC = 8
$$

---

### Step 3: Construct the Cubic Equation

We now know that $A, B, C$ are the roots of the cubic equation:

$$
x^3 - (A + B + C)x^2 + (AB + BC + CA)x - ABC = x^3 - 0x^2 + 0x - 8 = x^3 - 8
$$

So the roots of this cubic are the cube roots of 8. These are:

$$
2, \quad -1 + i\sqrt{3}, \quad -1 - i\sqrt{3}
$$

---

### Final Answer

The unordered set of the three expressions is:

$$
\boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}
$$

[Assistant]
Explanation:
Okay, so I need to find the unordered set containing the three expressions: α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, and α₁α₄ + α₂α₃, where α₁, α₂, α₃, α₄ are the roots of the equation x⁴ + 2x³ + 2 = 0.

First, I remember that for polynomials, there are relationships between the coefficients and the roots, known as Vieta's formulas. Since this is a quartic equation, Vieta's formulas will relate the sums and products of the roots to the coefficients of the polynomial. Let me recall Vieta's formulas for a quartic equation.

For a general quartic equation x⁴ + a x³ + b x² + c x + d = 0, the sums are:

- α₁ + α₂ + α₃ + α₄ = -a
- α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = b
- α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -c
- α₁α₂α₃α₄ = d

But in our case, the equation is x⁴ + 2x³ + 2 = 0. Comparing with the general form, we have:

- a = 2, so the sum of roots α₁ + α₂ + α₃ + α₄ = -2
- b = 0 (since there is no x² term), so the sum of products two at a time is 0
- c = 0 (since there is no x term), so the sum of products three at a time is 0
- d = 2, so the product of all roots is 2

So, summarizing:

Sum of roots: S₁ = α₁ + α₂ + α₃ + α₄ = -2

Sum of products two at a time: S₂ = α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = 0

Sum of products three at a time: S₃ = α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = 0

Product of roots: S₄ = α₁α₂α₃α₄ = 2

Now, the problem asks for the set {α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃}. Let me denote these three expressions as follows:

Let’s call them:

A = α₁α₂ + α₃α₄

B = α₁α₃ + α₂α₄

C = α₁α₄ + α₂α₃

So, the set we need is {A, B, C}.

I need to find the values of A, B, C. Since the polynomial is quartic, and the roots are distinct, these expressions are symmetric in some way. Also, note that the sum of A, B, C is equal to S₂, which is 0. Because:

A + B + C = (α₁α₂ + α₃α₄) + (α₁α₃ + α₂α₄) + (α₁α₄ + α₂α₃) = (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄) = S₂ = 0.

Therefore, A + B + C = 0. So, the three numbers A, B, C add up to zero. Therefore, if I can find two of them, the third can be determined. But since the problem asks for the unordered set, maybe all three are equal? Or maybe they are different? Let me check.

Alternatively, perhaps there is a way to relate A, B, C with other symmetric sums. Let me think.

Alternatively, maybe I can consider that if I have the roots α₁, α₂, α₃, α₄, then perhaps I can pair them in different ways. However, since the polynomial is quartic, maybe there's a way to factor it or find relations between the roots?

Alternatively, perhaps I can consider that if I take two roots and multiply them, and then take the other two roots and multiply them, then add them. So, A = α₁α₂ + α₃α₄. Similarly for B and C.

But how can I find these values? Let me think. Since we know S₁, S₂, S₃, S₄, maybe we can find expressions for A, B, C. Let me think about how to relate them.

Let me note that if I consider the product of two roots, say α₁α₂, then perhaps I can relate it to other terms. But since we have symmetric sums, maybe we can find expressions for A, B, C in terms of S₁, S₂, S₃, S₄.

Alternatively, perhaps I can consider that if I take the product (α₁ + α₂)(α₃ + α₄). Let me compute that:

(α₁ + α₂)(α₃ + α₄) = α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ = B + C.

Similarly, if I take (α₁ + α₃)(α₂ + α₄) = α₁α₂ + α₁α₄ + α₃α₂ + α₃α₄ = A + C.

Similarly, (α₁ + α₄)(α₂ + α₃) = α₁α₂ + α₁α₃ + α₄α₂ + α₄α₃ = A + B.

So, if I denote:

Let me define:

Let’s say that:

Let’s denote p = α₁ + α₂, q = α₃ + α₄. Then, p + q = S₁ = -2.

Similarly, (α₁ + α₂)(α₃ + α₄) = p q = B + C.

Similarly, if I take other pairings, but maybe I need more relations.

Alternatively, perhaps I can find expressions for A, B, C. Let me think.

Alternatively, since A + B + C = 0, and if I can find AB + BC + CA, and ABC, then I can find the values of A, B, C as roots of a cubic equation. But how?

Alternatively, maybe I can find AB + BC + CA. Let me try.

First, let me compute AB + BC + CA. Let me compute AB + BC + CA.

But AB + BC + CA = A(B + C) + BC. Since A + B + C = 0, then B + C = -A. Therefore, AB + BC + CA = A(-A) + BC = -A² + BC. Similarly, but I need to find BC. Wait, but maybe this approach is getting too convoluted. Let me think again.

Alternatively, perhaps I can find expressions for A, B, C in terms of the known symmetric sums. Let me think.

Let me note that A = α₁α₂ + α₃α₄. Similarly, if I consider that the product of all roots is S₄ = 2. So, α₁α₂α₃α₄ = 2. Therefore, if I take A = α₁α₂ + α₃α₄, then if I multiply A by α₁α₂α₃α₄, but that might not help. Wait, maybe not.

Alternatively, perhaps I can consider that if I take the product of two roots, say α₁α₂, then the product of the other two roots is α₃α₄. Let me denote A = α₁α₂ + α₃α₄. Similarly, if I denote B = α₁α₃ + α₂α₄, and C = α₁α₄ + α₂α₃.

Alternatively, perhaps I can consider that if I take the square of the sum of roots:

(α₁ + α₂ + α₃ + α₄)² = α₁² + α₂² + α₃² + α₄² + 2(α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)

We know that S₁ = -2, so S₁² = 4. Also, S₂ = 0, so:

(-2)² = (sum of squares) + 2*(0) => 4 = sum of squares. Therefore, sum of squares of roots is 4.

But how does that help? Maybe not directly. Let me think.

Alternatively, perhaps I can consider the equations involving A, B, C. Let me think about the following:

Let me consider that if I take the product of two roots, say α₁α₂, then if I can relate it to other terms. For example, if I take α₁α₂, then since α₁ + α₂ + α₃ + α₄ = -2, then α₁ + α₂ = -2 - (α₃ + α₄). Let me denote p = α₁ + α₂, q = α₃ + α₄, so p + q = -2. Then, A = α₁α₂ + α₃α₄. Let me denote that as A = (α₁α₂) + (α₃α₄). Also, if I consider that α₁α₂ can be expressed in terms of p and something else. Wait, for example, if I know p = α₁ + α₂, then α₁α₂ can be found if I know something else. But how?

Alternatively, if I consider that α₁ and α₂ are roots of a quadratic equation, say x² - px + r = 0, where r = α₁α₂. Similarly, α₃ and α₄ are roots of another quadratic equation x² - qx + s = 0, where s = α₃α₄. Then, since the original quartic can be factored as (x² - px + r)(x² - qx + s) = 0. Let me try expanding this:

(x² - px + r)(x² - qx + s) = x⁴ - (p + q)x³ + (pq + r + s)x² - (ps + rq)x + rs.

Comparing with the original equation x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore, we can equate coefficients:

1. Coefficient of x⁴: 1 = 1, which is okay.

2. Coefficient of x³: -(p + q) = 2 => p + q = -2. Which matches our previous result since p + q = -2.

3. Coefficient of x²: pq + r + s = 0.

4. Coefficient of x: -(ps + rq) = 0 => ps + rq = 0.

5. Constant term: rs = 2.

So, we have these equations:

p + q = -2,

pq + r + s = 0,

ps + rq = 0,

rs = 2.

But we also know that A = r + s, since A = α₁α₂ + α₃α₄ = r + s. Similarly, if we can find r + s, which is A, and also other relations.

But we also have that p + q = -2, and pq + (r + s) = 0 => pq + A = 0 => pq = -A.

Also, from the fourth equation: ps + rq = 0. Let me note that p = α₁ + α₂, q = α₃ + α₄, r = α₁α₂, s = α₃α₄.

But how can I relate ps + rq? Let me think. Let me write ps + rq = p s + r q. Since p = α₁ + α₂, q = α₃ + α₄, r = α₁α₂, s = α₃α₄. So:

ps + rq = (α₁ + α₂)(α₃α₄) + (α₁α₂)(α₃ + α₄) = α₁α₃α₄ + α₂α₃α₄ + α₁α₂α₃ + α₁α₂α₄ = α₃α₄(α₁ + α₂) + α₁α₂(α₃ + α₄) = q s + p r. Wait, but that's the same as before. Wait, but we have ps + rq = 0. So, that's equal to q s + p r = 0.

But we also know that rs = 2. So, s = 2 / r. Therefore, substituting into the equation q s + p r = 0:

q*(2 / r) + p r = 0 => 2q / r + p r = 0 => Multiply both sides by r:

2q + p r² = 0 => p r² = -2q.

But we also have that pq = -A. So, p q = -A. Let me note that. Also, since p + q = -2, we can write q = -2 - p. Therefore, substituting into p q = -A:

p(-2 - p) = -A => -2p - p² = -A => A = 2p + p².

Similarly, from the equation p r² = -2q, and since q = -2 - p, then:

p r² = -2(-2 - p) = 4 + 2p => p r² = 4 + 2p => r² = (4 + 2p)/p = 4/p + 2.

But also, since rs = 2, and s = 2 / r, so s = 2 / r. Also, from pq + r + s = 0, which is pq + A = 0, since A = r + s. Wait, pq + A = 0 => pq = -A. But we already have that.

Alternatively, since A = r + s, and rs = 2, then r and s are roots of the quadratic equation t² - A t + 2 = 0. Therefore, r and s satisfy t² - A t + 2 = 0. Therefore, their product is 2, and their sum is A.

But also, from earlier, we have that r² = 4/p + 2. But since r is a root of t² - A t + 2 = 0, then r² = A r - 2. Therefore, equating the two expressions for r²:

A r - 2 = 4/p + 2 => A r = 4/p + 4 => A r = 4(1/p + 1) => A r = 4( (1 + p)/p )

But since p + q = -2, and q = -2 - p, and we also have that pq = -A. Let me recall that pq = -A. So, p(-2 - p) = -A => -2p - p² = -A => A = 2p + p². Therefore, A = p² + 2p.

So, substituting A into the previous equation:

A r = 4( (1 + p)/p )

=> (p² + 2p) r = 4( (1 + p)/p )

But since r is a root of t² - A t + 2 = 0, which is t² - (p² + 2p) t + 2 = 0. Therefore, r² = (p² + 2p) r - 2. But we also have that r² = 4/p + 2. Therefore, equating:

(p² + 2p) r - 2 = 4/p + 2 => (p² + 2p) r = 4/p + 4 => same as before.

But maybe I can express r in terms of p? Let me think. Since r is a root of t² - A t + 2 = 0, and A = p² + 2p, then:

t² - (p² + 2p) t + 2 = 0.

But also, from rs = 2, so s = 2 / r. Also, since A = r + s = r + 2 / r. Therefore, A = r + 2 / r. Therefore, multiplying both sides by r:

A r = r² + 2. But from the quadratic equation, r² = A r - 2. Therefore, substituting into A r = r² + 2 gives A r = (A r - 2) + 2 => A r = A r. Which is an identity. So, not helpful.

Hmm. Maybe I need another approach. Let me think again.

We have that A = p² + 2p, where p = α₁ + α₂. Also, since p + q = -2, q = -2 - p. Also, pq = -A. Therefore, p(-2 - p) = -A => A = 2p + p².

Additionally, we have that from the equation p r² = 4 + 2p. But since r² = A r - 2, substituting:

p(A r - 2) = 4 + 2p => p A r - 2p = 4 + 2p => p A r = 4 + 4p => A r = (4 + 4p)/p = 4(1 + p)/p.

But since A = p² + 2p, then:

(p² + 2p) r = 4(1 + p)/p => r = [4(1 + p)/p] / (p² + 2p) = [4(1 + p)/p] / [p(p + 2)] = 4(1 + p) / [p²(p + 2)].

But also, since A = r + 2 / r, we can write:

A = r + 2 / r. Let me substitute r from above:

A = [4(1 + p)/ (p²(p + 2))] + 2 / [4(1 + p)/ (p²(p + 2))]

Simplify the second term:

2 divided by [4(1 + p)/ (p²(p + 2))] = 2 * [p²(p + 2) / (4(1 + p))] = [p²(p + 2)] / [2(1 + p)]

Therefore, A = [4(1 + p)/ (p²(p + 2))] + [p²(p + 2)] / [2(1 + p)]

This seems complicated, but maybe we can combine these terms. Let me denote:

Let me write A as:

A = [4(1 + p)] / [p²(p + 2)] + [p²(p + 2)] / [2(1 + p)]

Let me find a common denominator. The denominators are p²(p + 2) and 2(1 + p). The common denominator would be 2p²(p + 2)(1 + p).

So, first term becomes:

[4(1 + p)] / [p²(p + 2)] * [2(1 + p)] / [2(1 + p)] = [8(1 + p)^2] / [2p²(p + 2)(1 + p)]

Wait, no. Wait, to get the common denominator, multiply numerator and denominator by 2(1 + p):

First term: [4(1 + p)] / [p²(p + 2)] * [2(1 + p)] / [2(1 + p)] = [4(1 + p) * 2(1 + p)] / [2p²(p + 2)(1 + p)] = [8(1 + p)^2] / [2p²(p + 2)(1 + p)] = [4(1 + p)] / [p²(p + 2)]

Wait, that seems like going in circles. Maybe another approach. Let me denote t = p. Then, A = t² + 2t. Also, from above, A = [4(1 + t)/ (t²(t + 2))] + [t²(t + 2)] / [2(1 + t)]. Let me compute this expression:

Let me compute the two terms:

Term1 = 4(1 + t)/[t²(t + 2)]

Term2 = [t²(t + 2)]/[2(1 + t)]

So, A = Term1 + Term2. But since A = t² + 2t, we can set up the equation:

t² + 2t = Term1 + Term2.

This seems complicated, but maybe we can substitute t² + 2t = A and see if we can find t. However, this seems too involved. Maybe there's a better way.

Alternatively, since we have that rs = 2 and A = r + s, then r and s are roots of t² - A t + 2 = 0. Also, from earlier, we have that p r² = 4 + 2p. But since r² = A r - 2, then:

p(A r - 2) = 4 + 2p => p A r - 2p = 4 + 2p => p A r = 4 + 4p => A r = (4 + 4p)/p = 4(1 + p)/p.

But since A = t² + 2t where t = p, and A = r + s, and r is a root of t² - A t + 2 = 0, then r can be expressed as [A ± sqrt(A² - 8)] / 2. But this seems too abstract.

Alternatively, maybe I can consider that since A = r + s and rs = 2, then A is the sum of two numbers whose product is 2. Therefore, A can be any real or complex number, but given that the original equation has complex roots, A could be complex. However, since the problem asks for the unordered set {A, B, C}, which are symmetric expressions, maybe they are all equal? Or maybe they are roots of some equation?

Alternatively, since A + B + C = 0, and if I can find AB + BC + CA and ABC, then I can find the cubic equation whose roots are A, B, C. Then, solving that equation would give me the values of A, B, C.

So, let me try to compute AB + BC + CA and ABC.

First, AB + BC + CA. Let me recall that:

AB + BC + CA = (A + B + C)^2 - (A² + B² + C²) all over 2? Wait, no. Wait, (A + B + C)^2 = A² + B² + C² + 2(AB + BC + CA). Therefore, AB + BC + CA = [(A + B + C)^2 - (A² + B² + C²)] / 2. But since A + B + C = 0, this simplifies to AB + BC + CA = - (A² + B² + C²)/2. But I don't know A² + B² + C².

Alternatively, maybe I can find AB + BC + CA directly. Let me think.

Let me note that A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me compute AB.

AB = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Let me expand this:

= α₁α₂α₁α₃ + α₁α₂α₂α₄ + α₃α₄α₁α₃ + α₃α₄α₂α₄

= α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²

Hmm, this seems complicated. Maybe there's a better way. Alternatively, perhaps I can use symmetric sums.

Alternatively, note that AB + BC + CA can be expressed in terms of the symmetric sums. Let me think.

Alternatively, consider that AB + BC + CA = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄) + (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃) + (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄).

This seems very complex, but maybe there's a pattern. Alternatively, perhaps I can use the fact that A + B + C = 0 and find AB + BC + CA by considering other relations.

Alternatively, let me think about the product of the roots. Since α₁α₂α₃α₄ = 2. Let me note that:

Let me consider that A = α₁α₂ + α₃α₄. Let me denote that as A. Similarly, if I multiply A by α₁α₂α₃α₄, which is 2, then:

A * 2 = (α₁α₂ + α₃α₄) * 2 = 2α₁α₂ + 2α₃α₄.

But not sure if helpful.

Alternatively, let me think about the following. Let me consider that if I take the product of all three expressions A, B, C. But that might be too complicated.

Alternatively, maybe I can use Newton's identities? But I'm not sure.

Alternatively, maybe I can consider that since the original equation is x⁴ + 2x³ + 2 = 0, maybe I can find relations between the roots. Let me try to factor the polynomial.

Let me try to factor x⁴ + 2x³ + 2. Maybe it factors into quadratics? Let me assume that it factors as (x² + a x + b)(x² + c x + d). Then expanding:

x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + b d.

Comparing with original equation x⁴ + 2x³ + 0x² + 0x + 2. Therefore, we have the following equations:

1. a + c = 2

2. ac + b + d = 0

3. ad + bc = 0

4. b d = 2

So, similar to earlier equations. Let me try to solve these equations.

From equation 1: c = 2 - a.

From equation 4: b d = 2. Let me assume that b and d are real numbers? Not necessarily, since the roots are complex. But maybe we can find integer solutions? Let me check.

Possible integer pairs for b and d such that b d = 2: (1, 2), (2, 1), (-1, -2), (-2, -1). Let me try these possibilities.

First, let me try b = 1, d = 2. Then equation 3: a*2 + b*c = 0 => 2a + 1*c = 0. But c = 2 - a, so 2a + (2 - a) = 0 => a + 2 = 0 => a = -2. Then c = 2 - (-2) = 4. Then check equation 2: ac + b + d = (-2)(4) + 1 + 2 = -8 + 3 = -5 ≠ 0. Not good.

Next, try b = 2, d = 1. Then equation 3: a*1 + 2*c = 0 => a + 2c = 0. Since c = 2 - a, substitute: a + 2(2 - a) = 0 => a + 4 - 2a = 0 => -a + 4 = 0 => a = 4. Then c = 2 - 4 = -2. Check equation 2: ac + b + d = (4)(-2) + 2 + 1 = -8 + 3 = -5 ≠ 0. Not good.

Next, try b = -1, d = -2. Equation 3: a*(-2) + (-1)*c = 0 => -2a - c = 0 => c = -2a. But c = 2 - a, so -2a = 2 - a => -2a + a = 2 => -a = 2 => a = -2. Then c = 2 - (-2) = 4. Check equation 2: ac + b + d = (-2)(4) + (-1) + (-2) = -8 - 3 = -11 ≠ 0. Not good.

Next, try b = -2, d = -1. Equation 3: a*(-1) + (-2)*c = 0 => -a - 2c = 0 => a = -2c. But c = 2 - a, so substitute: a = -2(2 - a) => a = -4 + 2a => -a = -4 => a = 4. Then c = 2 - 4 = -2. Check equation 2: ac + b + d = (4)(-2) + (-2) + (-1) = -8 - 3 = -11 ≠ 0. Not good.

So none of the integer pairs work. Maybe the factors are not with integer coefficients. Let me try to think differently.

Alternatively, maybe the quartic can be written as x⁴ + 2x³ + 2 = 0. Let me try to make a substitution. Let me set y = x + k, to eliminate the cubic term? Wait, but the cubic term is already present. Alternatively, maybe divide by x²? Let me try:

x⁴ + 2x³ + 2 = 0 => x² + 2x + 2/x² = 0. Let me set z = x + 1/x. Then z² = x² + 2 + 1/x². Therefore, x² + 1/x² = z² - 2. But our equation is x² + 2x + 2/x² = 0. Let me write it as (x² + 2/x²) + 2x = 0. But x² + 2/x² = (x² + 1/x²) + 1/x² = (z² - 2) + 1/x². Not sure. Alternatively, maybe not helpful.

Alternatively, maybe write the equation as x⁴ + 2x³ = -2. Then factor x³(x + 2) = -2. Not sure.

Alternatively, maybe consider that the equation is x⁴ + 2x³ + 2 = 0. Let me try to find roots numerically? But since the problem is about symmetric expressions, maybe not helpful.

Alternatively, maybe consider that since the polynomial is x⁴ + 2x³ + 2, and we know Vieta's formulas, maybe we can find relations between A, B, C.

Earlier, I had that A + B + C = 0. Also, if I can find AB + BC + CA and ABC, then I can find the cubic equation whose roots are A, B, C.

Let me try to compute AB + BC + CA.

Let me recall that A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me compute AB + BC + CA.

First, compute AB:

AB = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄)

= α₁α₂α₁α₃ + α₁α₂α₂α₄ + α₃α₄α₁α₃ + α₃α₄α₂α₄

= α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²

Similarly, compute BC:

BC = (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃)

= α₁α₃α₁α₄ + α₁α₃α₂α₃ + α₂α₄α₁α₄ + α₂α₄α₂α₃

= α₁²α₃α₄ + α₁α₂α₃² + α₁α₂α₄² + α₂²α₃α₄

Similarly, compute CA:

CA = (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄)

= α₁α₄α₁α₂ + α₁α₄α₃α₄ + α₂α₃α₁α₂ + α₂α₃α₃α₄

= α₁²α₂α₄ + α₁α₃α₄² + α₁α₂²α₃ + α₂α₃²α₄

Now, adding AB + BC + CA:

Let me list all terms from AB, BC, CA:

From AB:

1. α₁²α₂α₃

2. α₁α₂²α₄

3. α₁α₃²α₄

4. α₂α₃α₄²

From BC:

5. α₁²α₃α₄

6. α₁α₂α₃²

7. α₁α₂α₄²

8. α₂²α₃α₄

From CA:

9. α₁²α₂α₄

10. α₁α₃α₄²

11. α₁α₂²α₃

12. α₂α₃²α₄

Now, let me group similar terms:

Terms with α₁²α₂α₃: terms 1 and 11. Term 1 is α₁²α₂α₃, term 11 is α₁α₂²α₃. Wait, no, term 11 is α₁α₂²α₃? Wait, term 11 is from CA: α₁α₂²α₃? Let me check:

From CA: term 11 is α₁α₂²α₃? Wait, no. Let me recheck:

From CA:

Term 9: α₁²α₂α₄

Term 10: α₁α₃α₄²

Term 11: α₁α₂²α₃

Term 12: α₂α₃²α₄

So, term 11 is α₁α₂²α₃, term 1 is α₁²α₂α₃. So, terms 1 and 11 are α₁²α₂α₃ and α₁α₂²α₃. Similarly, terms 2 and 6: term 2 is α₁α₂²α₄, term 6 is α₁α₂α₃². Wait, no. Let me check:

Wait, term 2 is α₁α₂²α₄, term 6 is α₁α₂α₃². Not the same.

This seems too messy. Maybe there's a better way. Let me think about the sum AB + BC + CA.

Alternatively, note that AB + BC + CA can be expressed as:

AB + BC + CA = (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄) + (α₁α₃ + α₂α₄)(α₁α₄ + α₂α₃) + (α₁α₄ + α₂α₃)(α₁α₂ + α₃α₄)

Let me denote that each term is a product of two of the expressions A, B, C. But since A + B + C = 0, maybe there's a relation.

Alternatively, consider that AB + BC + CA = (A + B + C)^2 - (A² + B² + C²) all over 2, but since A + B + C = 0, this becomes - (A² + B² + C²)/2. But I don't know A² + B² + C².

Alternatively, maybe I can find A² + B² + C². Let me try:

A² + B² + C² = (A + B + C)^2 - 2(AB + BC + CA) = 0 - 2(AB + BC + CA) = -2(AB + BC + CA).

But this seems circular.

Alternatively, maybe I can find A² + B² + C² by expanding A², B², C².

Let me compute A² = (α₁α₂ + α₃α₄)^2 = α₁²α₂² + 2α₁α₂α₃α₄ + α₃²α₄². Similarly for B² and C².

So, A² + B² + C² = (α₁²α₂² + α₃²α₄²) + (α₁²α₃² + α₂²α₄²) + (α₁²α₄² + α₂²α₃²) + 6α₁α₂α₃α₄.

Wait, because each A², B², C² has a term 2α₁α₂α₃α₄, so total of 3*2α₁α₂α₃α₄ = 6α₁α₂α₃α₄.

So, A² + B² + C² = [α₁²α₂² + α₃²α₄² + α₁²α₃² + α₂²α₄² + α₁²α₄² + α₂²α₃²] + 6α₁α₂α₃α₄.

Now, note that α₁²α₂² + α₁²α₃² + α₁²α₄² + α₂²α₃² + α₂²α₄² + α₃²α₄² = (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 - 2(α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄)(α₁ + α₂ + α₃ + α₄) + 2α₁α₂α₃α₄. Wait, this seems too complicated.

Alternatively, recall that (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = S₂² = 0² = 0.

But expanding this square gives:

(α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = Σα_i²α_j² + 2Σα_iα_jα_kα_l where i < j, k < l. Wait, no. Wait, more carefully:

When you square the sum of products two at a time, you get the sum of squares of each product plus twice the sum of products of distinct pairs of products.

So, Σ(α_iα_j)^2 + 2Σ_{i < j, k < l, (i,j) ≠ (k,l)} α_iα_jα_kα_l.

But this seems complex. However, we know that Σ(α_iα_j)^2 = A² + B² + C² - 6α₁α₂α₃α₄? Wait, no. Earlier, we had:

A² + B² + C² = Σ(α_iα_j)^2 + 6α₁α₂α₃α₄.

Wait, earlier, I had:

A² + B² + C² = [Σ(α_iα_j)^2] + 6α₁α₂α₃α₄.

But since S₂ = Σ(α_iα_j) = 0, then S₂² = [Σ(α_iα_j)]² = 0 = Σ(α_iα_j)^2 + 2Σ_{i < j, k < l} α_iα_jα_kα_l.

But Σ_{i < j, k < l} α_iα_jα_kα_l is equal to the sum over all products of two distinct pairs. However, this is equal to (Σα_iα_j)(Σα_kα_l) - Σ(α_iα_j)^2. Wait, no. Wait, if I consider (Σα_iα_j)(Σα_kα_l) = Σα_iα_jα_kα_l + Σα_iα_jα_kα_l where the pairs overlap. Wait, this is getting too tangled.

Alternatively, since we know that Σ(α_iα_j)^2 = S₂² - 2Σα_iα_jα_kα_l, but I'm not sure.

Alternatively, since we know that α₁α₂α₃α₄ = 2, so the product of all roots is 2. Let me think about Σ(α_iα_j)^2.

Let me note that Σ(α_iα_j)^2 = (α₁α₂)^2 + (α₁α₃)^2 + (α₁α₄)^2 + (α₂α₃)^2 + (α₂α₄)^2 + (α₃α₄)^2.

Let me consider that (α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄)^2 = 0 = Σ(α_iα_j)^2 + 2Σ_{i < j, k < l} α_iα_jα_kα_l.

But Σ_{i < j, k < l} α_iα_jα_kα_l is equal to the sum over all products of two distinct pairs. How many terms are there? For four variables, the number of ways to choose two distinct pairs is C(6, 2) minus the cases where the pairs overlap. Wait, no. Actually, the number of ways to choose two distinct pairs is C(6, 2), but many of these pairs overlap.

Alternatively, note that Σ_{i < j, k < l} α_iα_jα_kα_l = (Σα_iα_j)(Σα_kα_l) - Σ(α_iα_j)^2. Wait, no. Wait, if I multiply Σα_iα_j by Σα_kα_l, I get Σα_iα_jα_kα_l + Σα_iα_jα_kα_l where the pairs (i,j) and (k,l) overlap. But this is too vague.

Alternatively, since we know that Σα_iα_j = 0, and Σα_iα_jα_k = 0, and Σα_iα_jα_kα_l = 2.

Wait, let me think. The product of all four roots is 2. So, if I take the product of two pairs, say α₁α₂α₃α₄ = 2. So, for any two pairs, their product is 2. For example, (α₁α₂)(α₃α₄) = 2, (α₁α₃)(α₂α₄) = 2, (α₁α₄)(α₂α₃) = 2. Therefore, each of the products of two pairs is 2.

Therefore, Σ_{i < j, k < l} α_iα_jα_kα_l = 3 * 2 = 6? Wait, no. Wait, there are three ways to split the four roots into two pairs: (α₁α₂)(α₃α₄), (α₁α₃)(α₂α₄), (α₁α₄)(α₂α₃). Each of these products is equal to 2. Therefore, the sum of all such products is 3 * 2 = 6.

But earlier, we had that (Σα_iα_j)^2 = Σ(α_iα_j)^2 + 2 * Σ_{i < j, k < l} α_iα_jα_kα_l.

Since Σα_iα_j = 0, then 0 = Σ(α_iα_j)^2 + 2 * 6 => Σ(α_iα_j)^2 = -12.

Wait, this is interesting! So, Σ(α_iα_j)^2 = -12.

Therefore, going back to A² + B² + C² = Σ(α_iα_j)^2 + 6α₁α₂α₃α₄ = -12 + 6*2 = -12 + 12 = 0.

So, A² + B² + C² = 0.

But earlier, we had that AB + BC + CA = - (A² + B² + C²)/2 = -0/2 = 0.

So, AB + BC + CA = 0.

Therefore, now we know that A + B + C = 0, AB + BC + CA = 0, and we need to find ABC.

If we can find ABC, then we can write the cubic equation whose roots are A, B, C as x³ - (A + B + C)x² + (AB + BC + CA)x - ABC = x³ - 0x² + 0x - ABC = x³ - ABC = 0. Therefore, the cubic equation is x³ - ABC = 0, so the roots are the cube roots of ABC.

But since A, B, C are complex numbers, and the original equation has complex roots, it's possible that ABC is a real number or complex. Let me try to find ABC.

Let me compute ABC.

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

This seems complicated, but maybe there's a way to simplify it. Let me think.

Let me recall that earlier, we had that each pair product is 2. For example, (α₁α₂)(α₃α₄) = 2, (α₁α₃)(α₂α₄) = 2, (α₁α₄)(α₂α₃) = 2.

Let me denote that:

Let me denote A = α₁α₂ + α₃α₄, B = α₁α₃ + α₂α₄, C = α₁α₄ + α₂α₃.

Let me consider multiplying A, B, C.

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

Let me first compute (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Earlier, we expanded this and got terms like α₁²α₂α₃ + ... but maybe there's a smarter way.

Alternatively, let me note that if I let x = α₁α₂, y = α₃α₄, z = α₁α₃, w = α₂α₄, v = α₁α₄, u = α₂α₃. Then, A = x + y, B = z + w, C = v + u. But this might not help.

Alternatively, let me consider that since (α₁α₂)(α₃α₄) = 2, so x y = 2, similarly z w = 2, v u = 2.

But how does this help?

Alternatively, let me consider that ABC = (A)(B)(C). Let me try to compute this product.

Let me first compute (α₁α₂ + α₃α₄)(α₁α₃ + α₂α₄). Let me denote this as (x + y)(z + w), where x = α₁α₂, y = α₃α₄, z = α₁α₃, w = α₂α₄.

Expanding this gives xz + xw + yz + yw.

Now, xz = α₁α₂ * α₁α₃ = α₁²α₂α₃.

xw = α₁α₂ * α₂α₄ = α₁α₂²α₄.

yz = α₃α₄ * α₁α₃ = α₁α₃²α₄.

yw = α₃α₄ * α₂α₄ = α₂α₃α₄².

So, the product (x + y)(z + w) = α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄².

Now, multiply this by (v + u) = α₁α₄ + α₂α₃. Let me denote this as (v + u).

So, ABC = [α₁²α₂α₃ + α₁α₂²α₄ + α₁α₃²α₄ + α₂α₃α₄²] * (α₁α₄ + α₂α₃).

This seems very complex, but maybe we can find a pattern or use known values.

Let me note that α₁α₂α₃α₄ = 2. So, for example, α₁α₂α₃ = 2 / α₄, and similarly for other terms. Let me try to express each term in the first bracket in terms of α₁α₂α₃α₄.

For example, α₁²α₂α₃ = α₁ * (α₁α₂α₃) = α₁ * (2 / α₄). Similarly, α₁α₂²α₄ = α₂ * (α₁α₂α₄) = α₂ * (2 / α₃). Similarly, α₁α₃²α₄ = α₃ * (α₁α₃α₄) = α₃ * (2 / α₂). And α₂α₃α₄² = α₄ * (α₂α₃α₄) = α₄ * (2 / α₁).

Therefore, the first bracket becomes:

α₁*(2 / α₄) + α₂*(2 / α₃) + α₃*(2 / α₂) + α₄*(2 / α₁) = 2[α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁].

Now, multiplying this by (α₁α₄ + α₂α₃):

ABC = 2[α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁] * (α₁α₄ + α₂α₃).

Let me compute this product. Let me denote the first bracket as 2K, where K = α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁.

So, ABC = 2K * (α₁α₄ + α₂α₃).

Let me compute K * (α₁α₄ + α₂α₃).

Let me expand this:

K * (α₁α₄ + α₂α₃) = [α₁/α₄ + α₂/α₃ + α₃/α₂ + α₄/α₁] * (α₁α₄ + α₂α₃).

Let me distribute each term:

First term: α₁/α₄ * α₁α₄ = α₁²

Second term: α₁/α₄ * α₂α₃ = α₁α₂α₃ / α₄

Third term: α₂/α₃ * α₁α₄ = α₁α₂α₄ / α₃

Fourth term: α₂/α₃ * α₂α₃ = α₂²

Fifth term: α₃/α₂ * α₁α₄ = α₁α₃α₄ / α₂

Sixth term: α₃/α₂ * α₂α₃ = α₃²

Seventh term: α₄/α₁ * α₁α₄ = α₄²

Eighth term: α₄/α₁ * α₂α₃ = α₂α₃α₄ / α₁

So, combining all terms:

α₁² + α₂² + α₃² + α₄² + [α₁α₂α₃ / α₄ + α₁α₂α₄ / α₃ + α₁α₃α₄ / α₂ + α₂α₃α₄ / α₁]

Now, let me simplify each of the bracketed terms.

Note that α₁α₂α₃ / α₄ = (α₁α₂α₃α₄) / α₄² = 2 / α₄². Similarly:

α₁α₂α₃ / α₄ = 2 / α₄²

α₁α₂α₄ / α₃ = 2 / α₃²

α₁α₃α₄ / α₂ = 2 / α₂²

α₂α₃α₄ / α₁ = 2 / α₁²

Therefore, the bracketed terms become 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

So, overall, K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Therefore, ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²) ].

Now, we need to compute α₁² + α₂² + α₃² + α₄² and 1/α₁² + 1/α₂² + 1/α₃² + 1/α₄².

Earlier, we found that the sum of squares of the roots is 4. Let me recall that:

From earlier, we had that (α₁ + α₂ + α₃ + α₄)^2 = 4 = sum of squares + 2 * sum of products two at a time. Since sum of products two at a time is 0, then sum of squares is 4. So, α₁² + α₂² + α₃² + α₄² = 4.

Now, let me compute 1/α₁² + 1/α₂² + 1/α₃² + 1/α₄².

Let me denote this as S = Σ1/α_i².

Note that Σ1/α_i² = (Σ1/α_i)^2 - 2Σ1/(α_iα_j).

Let me compute Σ1/α_i and Σ1/(α_iα_j).

First, Σ1/α_i = (α₂α₃α₄ + α₁α₃α₄ + α₁α₂α₄ + α₁α₂α₃) / (α₁α₂α₃α₄).

From Vieta's formulas, the sum of products three at a time is S₃ = 0, and the product of all roots is S₄ = 2. Therefore, Σ1/α_i = S₃ / S₄ = 0 / 2 = 0.

Next, Σ1/(α_iα_j) = Σ(α_kα_l) / (α_iα_jα_kα_l) where α_kα_l is the product of the other two roots. Wait, let me think.

Σ1/(α_iα_j) = Σ_{i < j} 1/(α_iα_j).

But since α₁α₂α₃α₄ = 2, then 1/(α_iα_j) = α_kα_l / 2, where α_kα_l is the product of the other two roots.

Therefore, Σ1/(α_iα_j) = (α₃α₄ + α₁α₄ + α₁α₃ + α₂α₃ + α₂α₄ + α₁α₂)/2 = S₂ / 2 = 0 / 2 = 0.

Therefore, Σ1/α_i² = (Σ1/α_i)^2 - 2Σ1/(α_iα_j) = 0² - 2*0 = 0.

Wait, that's interesting! So, Σ1/α_i² = 0.

Therefore, going back to ABC:

ABC = 2 * [4 + 2*0] = 2 * 4 = 8.

Wait, so ABC = 8?

Wait, let me check again:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ] = 2 * [4 + 2*0] = 2 * 4 = 8.

Yes, that seems correct.

Therefore, ABC = 8.

So, now we have that the cubic equation with roots A, B, C is x³ - ABC = x³ - 8 = 0. Therefore, the roots are the cube roots of 8.

But 8 is 2³, so the cube roots of 8 are 2, 2ω, 2ω², where ω is a primitive third root of unity.

But wait, the cubic equation is x³ - 8 = 0, which factors as (x - 2)(x² + 2x + 4) = 0. Therefore, the roots are 2, -1 + i√3, -1 - i√3.

But the problem says that the roots α₁, α₂, α₃, α₄ are distinct. So, the set {A, B, C} is the set containing 2, -1 + i√3, -1 - i√3.

But the problem asks for the unordered set {A, B, C}. So, the answer is {2, -1 + i√3, -1 - i√3}.

But let me check if this makes sense.

Wait, but earlier we had that A + B + C = 0, and the roots of x³ - 8 = 0 are 2, -1 ± i√3. Let me check if their sum is 0: 2 + (-1 + i√3) + (-1 - i√3) = 2 - 1 - 1 + i√3 - i√3 = 0. Yes, that works.

Also, AB + BC + CA = 0, as we found earlier. Let me check with these roots. Let me compute AB + BC + CA for roots 2, -1 + i√3, -1 - i√3.

Let me denote the roots as 2, ω, ω², where ω = -1 + i√3 and ω² = -1 - i√3.

Then AB + BC + CA = ω² + ω + 2ω² + 2ω + ω * 2? Wait, no. Wait, AB + BC + CA for the roots 2, ω, ω² is:

Let me compute AB + BC + CA where A = 2, B = ω, C = ω².

AB + BC + CA = 2ω + ω * ω² + ω² * 2 = 2ω + ω³ + 2ω².

But ω³ = 1, since ω is a primitive third root of unity. Therefore, AB + BC + CA = 2ω + 1 + 2ω².

But since ω + ω² = -1, then 2ω + 2ω² = 2(ω + ω²) = -2. Therefore, AB + BC + CA = -2 + 1 = -1. Wait, but earlier we had AB + BC + CA = 0. This is a contradiction!

Wait, this suggests an error in my calculation.

Wait, earlier I said that AB + BC + CA = 0, but according to this, it's -1. Where did I go wrong?

Let me retrace. Earlier, I had that A² + B² + C² = 0, and since A + B + C = 0, then AB + BC + CA = - (A² + B² + C²)/2 = 0. But according to the roots of x³ - 8 = 0, AB + BC + CA is not zero. So, there must be a mistake in my calculation of ABC.

Let me check the calculation of ABC again.

Earlier, I had ABC = 8, but if the roots are 2, ω, ω², then ABC = 2 * ω * ω² = 2 * ω³ = 2 * 1 = 2. But earlier I said ABC = 8. So, there's a contradiction here.

This suggests that my calculation of ABC was wrong. Let me go back.

Let me re-examine the calculation of ABC.

Earlier, I had:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ] = 2 * [4 + 2*0] = 8.

But if the roots are 2, ω, ω², then ABC should be 2 * ω * ω² = 2 * 1 = 2. So, there's a discrepancy.

Where is the mistake? Let me check the steps again.

Starting from ABC = 2 * [ (sum of squares) + 2(sum of reciprocals squared) ] = 2 * [4 + 0] = 8.

But if ABC is 2, then this suggests that my calculation is wrong.

Let me go back to the step where I calculated ABC.

I had:

ABC = 2 * [ (α₁² + α₂² + α₃² + α₄²) + 2(Σ1/α_i²) ]

But let me check that step again.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Then ABC = 2 * [ ... ]

But let me check the expansion again.

When I expanded K * (α₁α₄ + α₂α₃), I got:

α₁² + α₂² + α₃² + α₄² + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But let me verify this.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = (α₁² + α₂² + α₃² + α₄²) + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But let me check with specific values. Suppose all α_i are real numbers, but in our case, they are complex. However, let me take a simple case where α₁ = α₂ = α₃ = α₄ = 1. Then, the original equation would be x⁴ + 2x³ + 2 = 0, which is not satisfied, but just for testing.

But let me not get bogged down. Let me think about the derivation again.

When I expanded K * (α₁α₄ + α₂α₃), I had:

α₁² + α₂² + α₃² + α₄² + 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

But how did I get that? Let me re-examine the expansion.

Earlier, I had:

K * (α₁α₄ + α₂α₃) = α₁² + α₂² + α₃² + α₄² + [α₁α₂α₃ / α₄ + α₁α₂α₄ / α₃ + α₁α₃α₄ / α₂ + α₂α₃α₄ / α₁].

Then, I said that each of those terms is 2 / α_i².

Wait, let me check that.

For example, α₁α₂α₃ / α₄ = (α₁α₂α₃α₄) / α₄² = 2 / α₄². Yes, that's correct. Similarly for the others.

Therefore, the bracketed terms are 2(1/α₁² + 1/α₂² + 1/α₃² + 1/α₄²).

Therefore, K * (α₁α₄ + α₂α₃) = (sum of squares) + 2(sum of reciprocals squared).

Then, ABC = 2 * [sum of squares + 2(sum of reciprocals squared)] = 2 * [4 + 2*0] = 8.

But if ABC is 8, then the cubic equation is x³ - 8 = 0, which has roots 2, -1 ± i√3. But then AB + BC + CA should be 0, but according to the roots, it's -1.

Wait, let me calculate AB + BC + CA for the roots 2, -1 + i√3, -1 - i√3.

Let me denote A = 2, B = -1 + i√3, C = -1 - i√3.

Then AB + BC + CA = (2)(-1 + i√3) + (-1 + i√3)(-1 - i√3) + (-1 - i√3)(2).

Compute each term:

First term: 2*(-1 + i√3) = -2 + 2i√3

Second term: (-1 + i√3)(-1 - i√3) = (-1)(-1) + (-1)(-i√3) + (i√3)(-1) + (i√3)(-i√3) = 1 + i√3 - i√3 - i²*3 = 1 - (-1)*3 = 1 + 3 = 4

Third term: (-1 - i√3)*2 = -2 - 2i√3

Adding all terms: (-2 + 2i√3) + 4 + (-2 - 2i√3) = (-2 - 2 + 4) + (2i√3 - 2i√3) = 0 + 0 = 0.

Ah! So AB + BC + CA = 0, which matches our previous result. Earlier, I must have made a mistake in calculating AB + BC + CA. So, the roots are correct.

Therefore, the cubic equation is x³ - 8 = 0, with roots 2, -1 + i√3, -1 - i√3.

Therefore, the set {A, B, C} is {2, -1 + i√3, -1 - i√3}.

But the problem says "the unordered set", so we need to present it in boxed notation. However, the problem might expect the answer in a specific form. Let me check if there's a simpler way or if I made any miscalculations.

Alternatively, maybe there's a mistake in assuming that ABC = 8. Let me verify with another approach.

Let me recall that A, B, C are the three expressions, and we have A + B + C = 0, AB + BC + CA = 0, and ABC = 8. Therefore, the cubic equation is x³ - 8 = 0, which has roots as above.

Alternatively, since the original equation is x⁴ + 2x³ + 2 = 0, and we have the roots α₁, α₂, α₃, α₄, and we are to find the set {A, B, C}.

But let me think if there's another way to find A, B, C.

Alternatively, since we know that A = α₁α₂ + α₃α₄, and we know that α₁α₂α₃α₄ = 2, so α₁α₂ = 2 / (α₃α₄). Therefore, A = 2 / (α₃α₄) + α₃α₄. Let me denote y = α₃α₄, then A = y + 2/y. Similarly, since α₁α₂ = 2/y.

Similarly, if I consider that the product of α₁α₂ and α₃α₄ is 2, so they are reciprocal up to a constant.

But since A = y + 2/y, then A is a function of y. Similarly, if I consider other pairings, B and C would be similar expressions.

But since the roots are symmetric, the values of A, B, C should be the same as the roots of the equation t = y + 2/y, where y is a product of two roots.

But since the products of two roots are α₁α₂, α₁α₃, α₁α₄, α₂α₃, α₂α₄, α₃α₄. But we know that each of these products is either y or 2/y, depending on the pairing.

But since there are three distinct pairings (as A, B, C), each corresponding to different pairings, and each of these pairings would give a value of the form y + 2/y, where y is one of the products.

But since the products come in reciprocal pairs, each value of y + 2/y would be the same for reciprocal pairs.

But since there are three distinct values of A, B, C, each corresponding to different pairings, and each of these values satisfies t = y + 2/y, where y is a product of two roots.

But since the products of two roots are related to the roots of the equation.

Alternatively, since we have already derived that the set {A, B, C} is {2, -1 + i√3, -1 - i√3}, which are the roots of x³ - 8 = 0, then this must be the correct answer.

Therefore, the unordered set is {2, -1 + i√3, -1 - i√3}.

But let me check if this makes sense with the original equation.

Let me consider that if A = 2, then α₁α₂ + α₃α₄ = 2. Since α₁α₂α₃α₄ = 2, then α₁α₂ = 2 / α₃α₄. So, if α₁α₂ + α₃α₄ = 2, then 2 / α₃α₄ + α₃α₄ = 2. Let me set z = α₃α₄, then 2/z + z = 2 => z² - 2z + 2 = 0 => z = [2 ± √(4 - 8)] / 2 = [2 ± √(-4)] / 2 = 1 ± i.

So, α₃α₄ = 1 + i or 1 - i. Similarly, the other pairings would give similar results.

But since the original equation has complex roots, this seems plausible.

Therefore, the set {A, B, C} is {2, -1 + i√3, -1 - i√3}.

But let me check if these values are consistent with the original equation.

Alternatively, since the problem asks for the unordered set, and the answer is three complex numbers, which are 2, and the two complex conjugates.

Therefore, the final answer is the set containing 2, -1 + i√3, and -1 - i√3.

But let me check if there's a simpler way to write this. For example, the problem might expect the answer in terms of radicals or something else, but since it's an unordered set, and the answer is three elements, I think this is correct.

Therefore, the answer is \boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}.

**Final Answer**
\boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}

To solve the problem, we are given the quartic polynomial:

$$
x^4 + 2x^3 + 2 = 0
$$

Let the roots of this equation be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, which are distinct complex numbers. We are asked to determine the unordered set:

$$
\{\alpha_1\alpha_2 + \alpha_3\alpha_4, \alpha_1\alpha_3 + \alpha_2\alpha_4, \alpha_1\alpha_4 + \alpha_2\alpha_3\}
$$

---

### Step 1: Use Vieta's Formulas

From the polynomial $x^4 + 2x^3 + 2 = 0$, we can apply Vieta's formulas:

- Sum of roots: $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = -2$
- Sum of products of roots taken two at a time: $\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = 0$
- Sum of products of roots taken three at a time: $\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \alpha_1\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4 = 0$
- Product of all roots: $\alpha_1\alpha_2\alpha_3\alpha_4 = 2$

Let us denote the three expressions we are interested in as:

- $A = \alpha_1\alpha_2 + \alpha_3\alpha_4$
- $B = \alpha_1\alpha_3 + \alpha_2\alpha_4$
- $C = \alpha_1\alpha_4 + \alpha_2\alpha_3$

---

### Step 2: Use Symmetry and Vieta's Identities

We observe that:

$$
A + B + C = (\alpha_1\alpha_2 + \alpha_3\alpha_4) + (\alpha_1\alpha_3 + \alpha_2\alpha_4) + (\alpha_1\alpha_4 + \alpha_2\alpha_3) = \text{sum of all products of roots two at a time} = 0
$$

So:

$$
A + B + C = 0
$$

Next, we compute $AB + BC + CA$. This is a symmetric expression that can be derived using the identities of symmetric polynomials. After a detailed expansion and simplification (as shown in the thought process), we find:

$$
AB + BC + CA = 0
$$

Finally, we compute $ABC$. Through a careful expansion and substitution using Vieta's formulas, we find:

$$
ABC = 8
$$

---

### Step 3: Construct the Cubic Equation

We now know that $A, B, C$ are the roots of the cubic equation:

$$
x^3 - (A + B + C)x^2 + (AB + BC + CA)x - ABC = x^3 - 0x^2 + 0x - 8 = x^3 - 8
$$

So the roots of this cubic are the cube roots of 8. These are:

$$
2, \quad -1 + i\sqrt{3}, \quad -1 - i\sqrt{3}
$$

---

### Final Answer

The unordered set of the three expressions is:

$$
\boxed{\{-1 - i\sqrt{3}, -1 + i\sqrt{3}, 2\}}
$$

Solution:
Employing the elementary symmetric polynomials ($s_1 = \alpha_1+\alpha_2+\alpha_3+\alpha_4 = -2$, $s_2 = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = 0$, $s_3 = \alpha_1\alpha_2\alpha_3 + \alpha_2\alpha_3\alpha_4 + \alpha_3\alpha_4\alpha_1 + \alpha_4\alpha_1\alpha_2 = 0$, and $s_4 = \alpha_1\alpha_2\alpha_3\alpha_4 = 2$) we consider the polynomial \[
P(x) = (x-(\alpha_1\alpha_2+\alpha_3\alpha_4))(x-(\alpha_1\alpha_3+\alpha_2\alpha_4))(x-(\alpha_1\alpha_4+\alpha_2\alpha_3))
\]Because $P$ is symmetric with respect to $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials.  We compute \begin{eqnarray*}
P(x) & = & x^3 - s_2x^2 + (s_3s_1-4s_4)x + (-s_3^2-s_4s_1^2+s_4s_2) \\
& = & x^3 - 8x - 8 \\
& = & (x+2)(x^2-2x-4)
\end{eqnarray*}The roots of $P(x)$ are $-2$ and $1 \pm \sqrt{5}$, so the answer is $\boxed{\{1\pm\sqrt{5},-2\}}.$

$\textbf{Remarks:}$ It is easy to find the coefficients of $x^2$ and $x$ by expansion, and the constant term can be computed without the complete expansion and decomposition of $(\alpha_1\alpha_2+\alpha_3\alpha_4)(\alpha_1\alpha_3+\alpha_2\alpha_4)(\alpha_1\alpha_4+\alpha_2\alpha_3)$ by noting that the only nonzero 6th degree expressions in $s_1, s_2, s_3,$ and $s_4$ are $s_1^6$ and $s_4s_1^2$. The general polynomial $P$ constructed here is called the cubic resolvent and arises in Galois theory.