Okay, let me try to figure out how to solve this problem. So, we have three variables x, y, and z with three equations given. The goal is to find the value of xy + yz + zx. Hmm, let's see.

First, let me write down the equations again to make sure I have them right:

1. xyz = 4
2. x³ + y³ + z³ = 4
3. xy² + x²y + xz² + x²z + yz² + y²z = 12

And we need to find xy + yz + zx. Alright.

I remember that in problems involving symmetric sums, sometimes using symmetric identities or Vieta's formulas can help. Since we have three variables, maybe we can relate them to the roots of a cubic equation. Let me recall that if x, y, z are roots of a cubic equation, then:

t³ - (x + y + z)t² + (xy + yz + zx)t - xyz = 0

So, if I can find x + y + z, xy + yz + zx, and xyz, then I can express the cubic equation. But here, we already know xyz = 4. So, if we can find x + y + z and xy + yz + zx, we can write the equation. But we need to find xy + yz + zx. Hmm.

Alternatively, maybe there's a way to relate the given equations to these symmetric sums. Let's think.

The second equation is x³ + y³ + z³ = 4. There's a formula for x³ + y³ + z³ in terms of symmetric sums. The formula is:

x³ + y³ + z³ = (x + y + z)³ - 3(x + y + z)(xy + yz + zx) + 3xyz

So, if I can express x³ + y³ + z³ using this identity, then maybe I can plug in the known values. Let's try that.

Given that x³ + y³ + z³ = 4 and xyz = 4, let's substitute into the formula:

4 = (x + y + z)³ - 3(x + y + z)(xy + yz + zx) + 3*4

Simplify that:

4 = (x + y + z)³ - 3(x + y + z)(xy + yz + zx) + 12

Subtract 12 from both sides:

4 - 12 = (x + y + z)³ - 3(x + y + z)(xy + yz + zx)

Which gives:

-8 = (x + y + z)³ - 3(x + y + z)(xy + yz + zx)

Hmm, so that's one equation involving (x + y + z) and (xy + yz + zx). Let's denote S = x + y + z and P = xy + yz + zx. Then the equation becomes:

-8 = S³ - 3S P

So, equation (1): S³ - 3S P = -8

Okay, now we need another equation to relate S and P. Let's look at the third given equation:

xy² + x²y + xz² + x²z + yz² + y²z = 12

Hmm, this looks like a combination of terms. Let me see if I can factor or group them. Let's write them as:

xy² + x²y + xz² + x²z + yz² + y²z

I can group terms with common factors. For example, terms with x:

xy² + x²y + xz² + x²z = x(y² + x y + z² + x z)

Similarly, terms with y and z:

yz² + y²z = y z² + y² z = y z(z + y)

Wait, but maybe there's a better way. Let's see:

xy² + x²y = xy(y + x)

Similarly, xz² + x²z = xz(z + x)

And yz² + y²z = yz(z + y)

So, the entire expression is:

xy(x + y) + xz(x + z) + yz(y + z)

Yes, that's correct. So, we can write the third equation as:

xy(x + y) + xz(x + z) + yz(y + z) = 12

Now, let's see. If we denote:

(x + y + z) = S

Then, x + y = S - z, x + z = S - y, y + z = S - x

So, substituting back into the third equation:

xy(S - z) + xz(S - y) + yz(S - x) = 12

Let's expand this:

xy S - xy z + xz S - xz y + yz S - yz x = 12

Combine like terms:

S(xy + xz + yz) - 3xyz = 12

Because each term has a product of two variables, and there are three such terms, so 3xyz.

So, we have:

S P - 3xyz = 12

But we know that xyz = 4, so:

S P - 3*4 = 12

Simplify:

S P - 12 = 12

So, S P = 24

Therefore, equation (2): S P = 24

Now, we have two equations:

1. S³ - 3S P = -8
2. S P = 24

So, we can substitute equation (2) into equation (1). Since S P = 24, then 3S P = 72.

Therefore, equation (1) becomes:

S³ - 72 = -8

So, S³ = 64

Therefore, S = cube root of 64 = 4

So, S = x + y + z = 4

Now, from equation (2), S P = 24, and S = 4, so:

4 P = 24 => P = 6

Therefore, xy + yz + zx = P = 6

Wait, so that's the answer? Let me check through the steps again to make sure I didn't make a mistake.

First, we used the identity for x³ + y³ + z³, which gave us equation (1). Then, we manipulated the third equation to express it in terms of S and P, leading to equation (2). Substituted equation (2) into equation (1), solved for S, then found P. Seems correct.

Let me verify with the values. If S = 4, P = 6, and xyz = 4, let's check the second equation:

x³ + y³ + z³ = (x + y + z)^3 - 3(x + y + z)(xy + yz + zx) + 3xyz

Substitute S = 4, P = 6, xyz = 4:

4³ - 3*4*6 + 3*4 = 64 - 72 + 12 = 4, which matches the given equation. That's correct.

Third equation: xy² + x²y + xz² + x²z + yz² + y²z = 12

We expressed this as S P - 3xyz = 12

Substituting S P = 24, xyz = 4: 24 - 12 = 12, which is correct.

So, all equations check out. Therefore, the value of xy + yz + zx is 6.

**Final Answer**
\boxed{6}
Thus, the value of $xy + yz + zx$ is $\boxed{6}$.
</think>