Explanation:
Okay, so I need to find the minimum value of (x + y)(y + z) given that x, y, z are positive real numbers and satisfy the condition xyz(x + y + z) = 1. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is likely a problem from algebra or inequalities, maybe I can use AM-GM inequality or some substitution. Let me try to recall the AM-GM inequality. It states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply that here.

But before jumping into inequalities, let me write down what I need to minimize and the constraint:

Minimize (x + y)(y + z) subject to xyz(x + y + z) = 1.

Let me expand (x + y)(y + z) to see if that helps. Expanding it:

(x + y)(y + z) = xy + xz + y² + yz.

Hmm, not sure if that helps directly. Maybe I can think of variables substitution. Let me consider variables a, b, c or something else. Alternatively, maybe express variables in terms of each other.

Alternatively, since there are three variables and one constraint, maybe I can reduce the number of variables. Let me think. Let me try to fix some variables or set some relations.

Alternatively, maybe assume symmetry? But the expression (x + y)(y + z) isn't symmetric. It's symmetric in x and z if we swap them? Let me check:

If I swap x and z, then (x + y)(y + z) becomes (z + y)(y + x) which is the same. So it's symmetric in x and z. So maybe the minimum occurs when x = z? That might be a good assumption. Let me try that.

Assume x = z. Then, let me denote x = z = a, and y = b. Then, the constraint becomes:

xyz(x + y + z) = a * b * a * (a + b + a) = a²b(2a + b) = 1.

And the expression to minimize becomes (a + b)(b + a) = (a + b)². Wait, that's interesting. If x = z, then (x + y)(y + z) becomes (a + b)(b + a) = (a + b)^2. So we need to minimize (a + b)^2 given that a²b(2a + b) = 1.

So maybe this substitution simplifies the problem. Let me work with that. Let me denote S = a + b, and maybe express the constraint in terms of S and something else. But maybe not. Alternatively, let me set variables to make it easier. Let me set t = b/a, so that b = ta. Then, since a and b are positive, t is positive. Let me substitute into the constraint:

a² * (ta) * (2a + ta) = a² * ta * a(2 + t) = a^4 t (2 + t) = 1.

So, a^4 t (2 + t) = 1. Therefore, a^4 = 1 / [t(2 + t)].

Then, the expression to minimize is (a + b)^2 = (a + ta)^2 = a²(1 + t)^2.

So, substituting a² from above. Since a^4 = 1/[t(2 + t)], then a² = 1 / sqrt[t(2 + t)]? Wait, no. Wait, a^4 = 1/[t(2 + t)] implies that a² = [1 / (t(2 + t))]^{1/2} = 1 / sqrt[t(2 + t)].

Therefore, the expression (a + b)^2 = a²(1 + t)^2 = [1 / sqrt(t(2 + t))] * (1 + t)^2.

So, we need to minimize [ (1 + t)^2 ] / sqrt(t(2 + t)) for t > 0.

Let me denote f(t) = (1 + t)^2 / sqrt(t(2 + t)). Let me simplify this expression. Let me square it to make it easier? Maybe not. Alternatively, write it as:

f(t) = (1 + t)^2 / [sqrt(t) * sqrt(2 + t)].

Alternatively, write it as:

f(t) = (1 + t)^2 / [sqrt(t(2 + t))].

Let me make substitution u = t, so we can take derivative. Maybe calculus is needed here. Since this seems complicated for AM-GM. Let me try calculus.

Let me compute derivative of f(t). Let me denote f(t) = (1 + t)^2 / [sqrt(t(2 + t))]. Let me write it as f(t) = (1 + t)^2 * [t(2 + t)]^{-1/2}.

Let me take natural logarithm to make differentiation easier. Let me set ln f(t) = 2 ln(1 + t) - (1/2) ln[t(2 + t)].

Then, derivative of ln f(t) is [2/(1 + t)] - (1/2)[ (1/(t(2 + t))) * ( (2 + t) + t ) ].

Wait, let me compute derivative step by step. Let me denote:

Let me compute d/dt [ln f(t)] = d/dt [2 ln(1 + t) - (1/2) ln(t(2 + t))].

So derivative is 2*(1/(1 + t)) - (1/2)*[ derivative of ln(t(2 + t)) ].

Derivative of ln(t(2 + t)) is derivative of ln t + ln(2 + t) = 1/t + 1/(2 + t).

Therefore, derivative of ln f(t) is 2/(1 + t) - (1/2)[1/t + 1/(2 + t)].

Set derivative equal to zero for minimization.

So:

2/(1 + t) - (1/2)[1/t + 1/(2 + t)] = 0.

Multiply both sides by 2 to eliminate the fraction:

4/(1 + t) - [1/t + 1/(2 + t)] = 0.

So:

4/(1 + t) = 1/t + 1/(2 + t).

Let me compute the right-hand side:

1/t + 1/(2 + t) = [ (2 + t) + t ] / [t(2 + t)] = (2 + 2t)/[t(2 + t)].

Therefore, equation becomes:

4/(1 + t) = (2 + 2t)/[t(2 + t)].

Simplify numerator on RHS: 2(1 + t). Therefore:

4/(1 + t) = 2(1 + t)/[t(2 + t)].

Multiply both sides by (1 + t):

4 = 2(1 + t)^2 / [t(2 + t)].

Multiply both sides by t(2 + t):

4t(2 + t) = 2(1 + t)^2.

Divide both sides by 2:

2t(2 + t) = (1 + t)^2.

Expand both sides:

Left side: 4t + 2t².

Right side: 1 + 2t + t².

Bring all terms to left side:

4t + 2t² - 1 - 2t - t² = 0.

Simplify:

(2t² - t²) + (4t - 2t) - 1 = 0 => t² + 2t - 1 = 0.

Solve quadratic equation t² + 2t - 1 = 0.

Solutions are t = [-2 ± sqrt(4 + 4)] / 2 = [-2 ± sqrt(8)] / 2 = [-2 ± 2*sqrt(2)] / 2 = -1 ± sqrt(2).

Since t > 0, we take the positive root: t = -1 + sqrt(2) ≈ -1 + 1.414 ≈ 0.414. Wait, sqrt(2) is approx 1.414, so -1 + sqrt(2) ≈ 0.414, which is positive. The other root is negative, so we discard it.

Therefore, critical point at t = sqrt(2) - 1.

Now, need to check if this is a minimum. Since the function tends to infinity as t approaches 0+ and as t approaches infinity, this critical point is likely to be the minimum.

So, let me compute the value of f(t) at t = sqrt(2) - 1.

First, compute t = sqrt(2) - 1. Let me compute 1 + t = 1 + sqrt(2) - 1 = sqrt(2). Also, 2 + t = 2 + sqrt(2) - 1 = 1 + sqrt(2).

Therefore, f(t) = (1 + t)^2 / sqrt(t(2 + t)) = [ (sqrt(2))^2 ] / sqrt( [sqrt(2) - 1][1 + sqrt(2)] )

Compute numerator: (sqrt(2))² = 2.

Denominator: sqrt( [sqrt(2) - 1][sqrt(2) + 1] ). Note that [sqrt(2) - 1][sqrt(2) + 1] = (sqrt(2))² - 1² = 2 - 1 = 1. Therefore, denominator is sqrt(1) = 1.

Therefore, f(t) = 2 / 1 = 2. So the minimal value of (a + b)^2 is 2. Therefore, the minimal value of (x + y)(y + z) is 2? Wait, but hold on. Wait, we assumed that x = z. So, does this mean that the minimal value is 2? But let me check if this is indeed the global minimum or if there's a lower value when x ≠ z.

But before that, let me check my steps again. So, assuming x = z, we found that the minimal value is 2. But maybe there's a lower value when x ≠ z. How can I be sure?

Alternatively, maybe the minimal value is indeed 2, achieved when x = z and t = sqrt(2) - 1. Let me check with actual numbers.

Let me compute a and b. Recall that t = b/a = sqrt(2) - 1. So, b = a*(sqrt(2) - 1). Also, from earlier, a^4 * t*(2 + t) = 1. Let me compute t*(2 + t):

t = sqrt(2) - 1, so 2 + t = 2 + sqrt(2) - 1 = 1 + sqrt(2). Therefore, t*(2 + t) = (sqrt(2) - 1)(sqrt(2) + 1) = 2 - 1 = 1. Therefore, a^4 * 1 = 1 => a^4 = 1 => a = 1 (since a > 0). Therefore, a = 1, so b = sqrt(2) - 1. Therefore, x = z = 1, y = sqrt(2) - 1.

Let me check the original constraint: xyz(x + y + z) = 1 * (sqrt(2) - 1) * 1 * (1 + sqrt(2) - 1 + 1) = (sqrt(2) - 1) * (sqrt(2) + 1) = 2 - 1 = 1. Correct. So the constraint is satisfied.

Then, (x + y)(y + z) = (1 + sqrt(2) - 1)(sqrt(2) - 1 + 1) = (sqrt(2))(sqrt(2)) = 2. Correct. So that works. So in this case, the value is 2.

But is this the minimum? Let me try another approach without assuming x = z. Maybe using Lagrange multipliers.

Let me set up the Lagrangian. Let me denote the function to minimize as f(x, y, z) = (x + y)(y + z), and the constraint is g(x, y, z) = xyz(x + y + z) - 1 = 0.

The Lagrangian is L = (x + y)(y + z) - λ(xyz(x + y + z) - 1).

Taking partial derivatives:

∂L/∂x = (y + z) - λ[yz(x + y + z) + xyz(1)] = 0.

Wait, let me compute derivative of f with respect to x:

df/dx = derivative of (x + y)(y + z) with respect to x is (1)(y + z) + 0 = y + z.

Similarly, derivative of g with respect to x is yz(x + y + z) + xyz(1) = yz(x + y + z) + xyz = yz(x + y + z + x) = yz(2x + y + z). Wait, no:

Wait, derivative of g = xyz(x + y + z) with respect to x is:

Use product rule: derivative of xyz times (x + y + z) + xyz times derivative of (x + y + z).

So:

First term: derivative of xyz with respect to x is yz*(x + y + z) + xyz*(derivative of (x + y + z) with respect to x) ?

Wait, no. Wait, actually, the function is xyz*(x + y + z). So derivative with respect to x is:

First, derivative of xyz*(x + y + z) with respect to x:

Let me denote u = xyz and v = x + y + z. Then derivative is u’v + uv’.

Compute u’ = derivative of xyz with respect to x is yz*(x + y + z) + xyz*(derivative of (x + y + z) with respect to x)? Wait, no. Wait, u = xyz, so derivative of u with respect to x is yz*(1) + 0 + 0? Wait, no. Wait, u = x * y * z. So derivative with respect to x is yz * 1 + 0 + 0 = yz. Wait, no, that's only if y and z are constants. But in reality, y and z are variables, but when taking partial derivatives, we treat other variables as constants. Wait, in the context of Lagrange multipliers, we take partial derivatives with respect to each variable, keeping others constant. Therefore, derivative of u = xyz with respect to x is yz. Then, derivative of v = x + y + z with respect to x is 1. Therefore, derivative of g with respect to x is yz*(x + y + z) + xyz*(1) = yz(x + y + z) + xyz = yz(x + y + z + x) = yz(2x + y + z). Wait, no:

Wait, it's yz(x + y + z) + xyz. Let me factor:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No, that would be if we factor yz. Wait, no:

Wait, yz(x + y + z) + xyz = yz(x + y + z + x)? No, that's not. Let me compute:

yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x)? No, that's not. Let me compute:

Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, xyz is x * y * z. So, if I factor yz, then:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No, that would be yz(x + y + z) + xyz = yz(x + y + z) + x y z = yz(x + y + z + x) ? No, that's not. Wait, maybe I need to compute it as:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Let me compute:

Let me write it as yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me take yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, maybe I need to factor something else. Let me compute:

Let me take yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me think of it as yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, maybe I need to compute:

Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me take the first term yz(x + y + z) and the second term xyz. Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, if I factor yz from both terms, but the second term is xyz, which is x * yz. Therefore, it's yz(x + y + z) + x * yz = yz [ (x + y + z) + x ] = yz(2x + y + z). Ah, yes! So derivative of g with respect to x is yz(2x + y + z). Similarly, partial derivatives with respect to y and z.

Therefore, the partial derivatives of L:

∂L/∂x = (y + z) - λ yz(2x + y + z) = 0.

Similarly, ∂L/∂y = derivative of f with respect to y is (x + y)'*(y + z) + (x + y)*(y + z)' = (1)(y + z) + (x + y)(1) = y + z + x + y = x + 2y + z.

Then, derivative of g with respect to y is:

Similarly, derivative of g with respect to y is xz(2y + x + z). Wait, let me check:

Using similar logic as before, derivative of g with respect to y is:

First term: derivative of xyz*(x + y + z) with respect to y is xz*(x + y + z) + xyz*(1) = xz(x + y + z) + xyz = xz(x + y + z + y) = xz(x + 2y + z). Wait, similar to before. So:

∂g/∂y = xz(x + 2y + z).

Therefore, ∂L/∂y = (x + 2y + z) - λ xz(x + 2y + z) = 0.

Similarly, ∂L/∂z = derivative of f with respect to z is (x + y) * derivative of (y + z) with respect to z = (x + y)(1) = x + y.

Derivative of g with respect to z is xy(x + y + 2z). Therefore:

∂L/∂z = (x + y) - λ xy(x + y + 2z) = 0.

So, we have three equations:

1) (y + z) - λ yz(2x + y + z) = 0.

2) (x + 2y + z) - λ xz(x + 2y + z) = 0.

3) (x + y) - λ xy(x + y + 2z) = 0.

And the constraint equation:

4) xyz(x + y + z) = 1.

This seems complicated, but maybe we can find relations between variables. Let me see.

Let me denote S = x + y + z. Then, note that equations 1, 2, 3 involve terms like 2x + y + z, which is S + x, similarly for others.

Let me rewrite equations:

Equation 1: (y + z) = λ yz(S + x).

Equation 2: (x + 2y + z) = λ xz(S + y).

Equation 3: (x + y) = λ xy(S + z).

Note that S = x + y + z, so S + x = 2x + y + z, S + y = x + 2y + z, S + z = x + y + 2z.

Let me denote:

Equation 1: (S - x) = λ yz(S + x).

Equation 2: (S - z) = λ xz(S + y).

Equation 3: (S - y) = λ xy(S + z).

Wait, because S - x = y + z, S - z = x + y, S - y = x + z. Wait, no:

Wait, S = x + y + z. Therefore:

S - x = y + z,

S - y = x + z,

S - z = x + y.

Therefore, equations 1, 2, 3 can be rewritten as:

1) (S - x) = λ yz(S + x).

2) (S - z) = λ xz(S + y).

3) (S - y) = λ xy(S + z).

So, equations:

1) (S - x) / [yz(S + x)] = λ.

2) (S - z) / [xz(S + y)] = λ.

3) (S - y) / [xy(S + z)] = λ.

Therefore, all three expressions equal to λ. Therefore, we can set them equal to each other.

Set equation 1 equal to equation 2:

(S - x)/[yz(S + x)] = (S - z)/[xz(S + y)].

Cross multiplying:

(S - x) * xz(S + y) = (S - z) * yz(S + x).

Cancel z from both sides:

(S - x) * x(S + y) = (S - z) * y(S + x).

Similarly, divide both sides by x and y? Wait, let me write:

(S - x) * x(S + y) = (S - z) * y(S + x).

Let me rearrange:

[ (S - x) / (S + x) ] * x = [ (S - z) / (S + y) ] * y.

Hmm, not sure. Alternatively, maybe assume some symmetry. Earlier, we assumed x = z and found a solution. Let me check if that satisfies these equations.

Assume x = z. Let me denote x = z = a, y = b. Then S = a + b + a = 2a + b.

Equation 1: (S - x)/[yz(S + x)] = (2a + b - a)/[b * a (2a + b + a)] = (a + b)/[ab(3a + b)].

Equation 2: (S - z)/[xz(S + y)] = same as equation 1, since x = z. So equations 1 and 2 are equal. Equation 3: (S - y)/[xy(S + z)] = (2a + b - b)/[a * b (2a + b + a)] = (2a)/[ab(3a + b)].

So, equations 1 and 3:

Equation 1: (a + b)/[ab(3a + b)] = λ.

Equation 3: 2a/[ab(3a + b)] = λ.

Therefore, equate them:

(a + b)/[ab(3a + b)] = 2a/[ab(3a + b)].

Multiply both sides by ab(3a + b):

a + b = 2a => b = a.

Wait, but earlier when we assumed x = z, we had t = b/a = sqrt(2) - 1, which is not 1. So this suggests that if x = z, then equations 1 and 3 would require b = a. But in our previous case, we had b ≠ a. Therefore, there is inconsistency here. Wait, that suggests that my assumption of x = z might not satisfy the Lagrangian conditions unless b = a. But earlier, when I assumed x = z and found a solution, it didn't satisfy this? Wait, but in that case, we had x = z = 1, y = sqrt(2) - 1 ≈ 0.414, so b ≠ a. Therefore, there must be a mistake here.

Wait, let me check again. If x = z = a, y = b, then equation 1 and equation 3:

Equation 1: (S - x)/[yz(S + x)] = (2a + b - a)/[b * a (2a + b + a)] = (a + b)/[ab(3a + b)].

Equation 3: (S - y)/[xy(S + z)] = (2a + b - b)/[a * b (2a + b + a)] = (2a)/[ab(3a + b)].

So equate equation 1 and equation 3:

(a + b)/[ab(3a + b)] = 2a/[ab(3a + b)] => a + b = 2a => b = a. Therefore, unless b = a, equations 1 and 3 are not equal. Therefore, if we assume x = z, then in order to satisfy the Lagrangian conditions, we must have b = a. But in our previous case, we had b ≠ a. Therefore, there is a contradiction. Which suggests that my previous approach of assuming x = z might not lead to the true minimum? Or maybe I made a mistake in the Lagrangian?

Wait, but when I assumed x = z, I found a critical point for the reduced problem, but that critical point does not satisfy the Lagrangian conditions? That seems contradictory. Wait, but maybe because when I reduced the problem by assuming x = z, I might have lost some generality? Or perhaps there is a mistake in the Lagrangian approach?

Alternatively, maybe there are multiple critical points, and the one found by assuming x = z is not the one from the Lagrangian? Wait, but how?

Alternatively, maybe my mistake was in the Lagrangian equations. Let me check again.

Wait, when I set up the Lagrangian, I had:

∂L/∂x = (y + z) - λ yz(2x + y + z) = 0.

But earlier, when I assumed x = z and found a solution, let me check if that solution satisfies the Lagrangian equations.

Let me take x = z = 1, y = sqrt(2) - 1 ≈ 0.414. Then S = x + y + z = 1 + (sqrt(2) - 1) + 1 = 1 + sqrt(2).

Compute equation 1: (y + z) - λ yz(2x + y + z) = 0.

Compute y + z = (sqrt(2) - 1) + 1 = sqrt(2).

Compute 2x + y + z = 2*1 + (sqrt(2) - 1) + 1 = 2 + sqrt(2).

Compute yz = (sqrt(2) - 1)*1 = sqrt(2) - 1.

Therefore, equation 1: sqrt(2) - λ*(sqrt(2) - 1)*(2 + sqrt(2)) = 0.

Compute (sqrt(2) - 1)*(2 + sqrt(2)):

Multiply out: sqrt(2)*2 + sqrt(2)*sqrt(2) - 1*2 - 1*sqrt(2) = 2 sqrt(2) + 2 - 2 - sqrt(2) = (2 sqrt(2) - sqrt(2)) + (2 - 2) = sqrt(2) + 0 = sqrt(2).

Therefore, equation 1 becomes sqrt(2) - λ*sqrt(2) = 0 => sqrt(2)(1 - λ) = 0 => λ = 1.

Now check equation 3: (x + y) - λ xy(S + z) = 0.

Compute x + y = 1 + (sqrt(2) - 1) = sqrt(2).

Compute xy = 1*(sqrt(2) - 1) = sqrt(2) - 1.

Compute S + z = (1 + sqrt(2)) + 1 = 2 + sqrt(2).

Therefore, equation 3: sqrt(2) - λ*(sqrt(2) - 1)*(2 + sqrt(2)) = sqrt(2) - 1*(sqrt(2)) = sqrt(2) - sqrt(2) = 0. Which is correct. Similarly, equation 2 would be same as equation 1 because x = z. Therefore, equations 1, 2, 3 are satisfied with λ = 1. Therefore, my previous thought that there was inconsistency was wrong. The mistake was in my algebra when I tried to equate equations 1 and 3. When I assumed x = z, equations 1 and 3 are not necessarily equal unless b = a, but in reality, with the specific values, they are equal because of the specific relation between variables. Therefore, the solution found by assuming x = z is indeed a critical point of the Lagrangian. Therefore, that critical point is valid.

Therefore, the minimal value is indeed 2. However, to be thorough, I need to check if there are other critical points where x ≠ z which might give a lower value.

Alternatively, maybe the minimal value is indeed 2, achieved at x = z, and that is the global minimum. To confirm, let me try another approach.

Let me consider using AM-GM inequality on the original problem. Let me think.

We need to minimize (x + y)(y + z) given that xyz(x + y + z) = 1.

Let me denote A = x + y + z. Then, the constraint is xyzA = 1.

We need to minimize (x + y)(y + z). Let me note that (x + y)(y + z) = y(x + y + z) + xz = yA + xz.

Therefore, (x + y)(y + z) = yA + xz.

So, we need to minimize yA + xz, given that xyzA = 1.

Hmm, interesting. So, we have to minimize yA + xz with xyzA = 1.

Let me think about how to relate these terms. Maybe use AM-GM on yA and xz?

But yA and xz are two terms. Let me consider that. Let me denote term1 = yA, term2 = xz. Then, by AM-GM, term1 + term2 ≥ 2√(term1 * term2). Therefore:

yA + xz ≥ 2√(yA * xz) = 2√(xyzA) = 2√(1) = 2.

Therefore, (x + y)(y + z) ≥ 2. Equality holds when yA = xz.

Therefore, the minimal value is 2, achieved when yA = xz and xyzA = 1.

So, this shows that the minimal value is indeed 2, and equality occurs when yA = xz and xyzA = 1. Let me check if this is compatible with our previous solution.

In our previous solution, x = z = 1, y = sqrt(2) - 1, A = x + y + z = 1 + sqrt(2) - 1 + 1 = 1 + sqrt(2). Then yA = (sqrt(2) - 1)(1 + sqrt(2)) = 2 - 1 = 1. And xz = 1 * 1 = 1. Therefore, yA = xz = 1, which satisfies the equality condition. Therefore, this is indeed the case where equality holds. Therefore, the minimal value is 2.

Therefore, regardless of assuming x = z or using AM-GM, we arrive at the conclusion that the minimal value is 2. Therefore, the answer is 2.

But just to make sure, let me try another example. Suppose I take different values. Let me try x = 2, y = 1, z = ?

Compute constraint: xyz(x + y + z) = 2 * 1 * z * (2 + 1 + z) = 2z(3 + z) = 1. Solve for z:

2z(3 + z) = 1 => 2z² + 6z - 1 = 0 => z = [-6 ± sqrt(36 + 8)] / 4 = [-6 ± sqrt(44)] / 4 = [-6 ± 2*sqrt(11)] / 4. Taking positive root: z = [ -6 + 2*sqrt(11) ] / 4 ≈ [ -6 + 6.633 ] / 4 ≈ 0.633 / 4 ≈ 0.158. Then compute (x + y)(y + z) = (2 + 1)(1 + 0.158) ≈ 3 * 1.158 ≈ 3.474, which is greater than 2. So indeed, higher than 2.

Another test case: take x = y = z. Let me see. Let x = y = z = t. Then constraint: t * t * t * (3t) = 3t^4 = 1 => t^4 = 1/3 => t = (1/3)^{1/4} ≈ 0.7598. Then (x + y)(y + z) = (2t)(2t) = 4t² ≈ 4*(1/3)^{1/2} ≈ 4 / 1.732 ≈ 2.309, which is still greater than 2. Therefore, the minimal value of 2 seems to hold.

Therefore, I think the minimal value is indeed 2, achieved when yA = xz and xyzA = 1, which occurs when x = z and y = sqrt(2) - 1, etc.

Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To find the **minimum value** of the expression
$$
(x + y)(y + z)
$$
under the constraint
$$
xyz(x + y + z) = 1,
$$
where $ x, y, z $ are **positive real numbers**, we proceed with a combination of **algebraic manipulation**, **substitution**, and **inequality techniques**.

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### Step 1: Expand the Expression

We begin by expanding the expression:
$$
(x + y)(y + z) = xy + xz + y^2 + yz.
$$
However, this form is not immediately helpful. Instead, we consider a more insightful approach.

---

### Step 2: Use Substitution and Symmetry

Let’s assume **symmetry** by setting $ x = z $. This is a reasonable assumption because the expression $ (x + y)(y + z) $ is symmetric in $ x $ and $ z $, and the constraint is also symmetric in $ x $ and $ z $.

Let $ x = z = a $, and $ y = b $. Then the constraint becomes:
$$
xyz(x + y + z) = a \cdot b \cdot a \cdot (a + b + a) = a^2 b (2a + b) = 1.
$$
The expression to minimize becomes:
$$
(x + y)(y + z) = (a + b)(b + a) = (a + b)^2.
$$

So, we now minimize:
$$
(a + b)^2 \quad \text{subject to} \quad a^2 b (2a + b) = 1.
$$

---

### Step 3: Use Substitution to Reduce Variables

Let’s define $ t = \frac{b}{a} $, so $ b = ta $. Then the constraint becomes:
$$
a^2 (ta) (2a + ta) = a^4 t (2 + t) = 1.
$$
So,
$$
a^4 = \frac{1}{t(2 + t)}.
$$
Then,
$$
a^2 = \frac{1}{\sqrt{t(2 + t)}}.
$$
Now, the expression to minimize becomes:
$$
(a + b)^2 = (a + ta)^2 = a^2 (1 + t)^2 = \frac{(1 + t)^2}{\sqrt{t(2 + t)}}.
$$

Let’s define:
$$
f(t) = \frac{(1 + t)^2}{\sqrt{t(2 + t)}}.
$$
We now minimize $ f(t) $ for $ t > 0 $.

---

### Step 4: Minimize the Function

Let’s compute the derivative of $ f(t) $ to find the critical points. However, instead of going through the full calculus, we can use a **substitution** to simplify the expression.

Let’s try $ t = \sqrt{2} - 1 $. Then:
- $ 1 + t = \sqrt{2} $,
- $ 2 + t = 1 + \sqrt{2} $,
- $ t(2 + t) = (\sqrt{2} - 1)(1 + \sqrt{2}) = 2 - 1 = 1 $.

So,
$$
f(t) = \frac{(\sqrt{2})^2}{\sqrt{1}} = \frac{2}{1} = 2.
$$

This is a **candidate for the minimum**.

---

### Step 5: Verify the Minimum

We now verify that this is indeed the **minimum** value. Consider the **AM-GM inequality**:

We can write:
$$
(x + y)(y + z) = y(x + y + z) + xz = yA + xz,
$$
where $ A = x + y + z $.

Then, by AM-GM:
$$
yA + xz \geq 2\sqrt{yA \cdot xz} = 2\sqrt{xyzA}.
$$
But from the constraint $ xyzA = 1 $, we get:
$$
(x + y)(y + z) \geq 2\sqrt{1} = 2.
$$
Equality holds when $ yA = xz $, which is satisfied in our earlier case.

---

### Step 6: Confirm the Equality Case

Let’s check if the equality condition is satisfied in our earlier solution:

Let $ x = z = 1 $, $ y = \sqrt{2} - 1 $, then:
- $ x + y + z = 1 + (\sqrt{2} - 1) + 1 = 1 + \sqrt{2} $,
- $ yA = (\sqrt{2} - 1)(1 + \sqrt{2}) = 2 - 1 = 1 $,
- $ xz = 1 \cdot 1 = 1 $,
- So $ yA = xz $, and the constraint is satisfied:
$$
xyz(x + y + z) = 1 \cdot (\sqrt{2} - 1) \cdot 1 \cdot (1 + \sqrt{2}) = 1.
$$

Thus, the minimum value is achieved, and the value is:
$$
(x + y)(y + z) = (1 + \sqrt{2} - 1)(\sqrt{2} - 1 + 1) = \sqrt{2} \cdot \sqrt{2} = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of (x + y)(y + z) given that x, y, z are positive real numbers and satisfy the condition xyz(x + y + z) = 1. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is likely a problem from algebra or inequalities, maybe I can use AM-GM inequality or some substitution. Let me try to recall the AM-GM inequality. It states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply that here.

But before jumping into inequalities, let me write down what I need to minimize and the constraint:

Minimize (x + y)(y + z) subject to xyz(x + y + z) = 1.

Let me expand (x + y)(y + z) to see if that helps. Expanding it:

(x + y)(y + z) = xy + xz + y² + yz.

Hmm, not sure if that helps directly. Maybe I can think of variables substitution. Let me consider variables a, b, c or something else. Alternatively, maybe express variables in terms of each other.

Alternatively, since there are three variables and one constraint, maybe I can reduce the number of variables. Let me think. Let me try to fix some variables or set some relations.

Alternatively, maybe assume symmetry? But the expression (x + y)(y + z) isn't symmetric. It's symmetric in x and z if we swap them? Let me check:

If I swap x and z, then (x + y)(y + z) becomes (z + y)(y + x) which is the same. So it's symmetric in x and z. So maybe the minimum occurs when x = z? That might be a good assumption. Let me try that.

Assume x = z. Then, let me denote x = z = a, and y = b. Then, the constraint becomes:

xyz(x + y + z) = a * b * a * (a + b + a) = a²b(2a + b) = 1.

And the expression to minimize becomes (a + b)(b + a) = (a + b)². Wait, that's interesting. If x = z, then (x + y)(y + z) becomes (a + b)(b + a) = (a + b)^2. So we need to minimize (a + b)^2 given that a²b(2a + b) = 1.

So maybe this substitution simplifies the problem. Let me work with that. Let me denote S = a + b, and maybe express the constraint in terms of S and something else. But maybe not. Alternatively, let me set variables to make it easier. Let me set t = b/a, so that b = ta. Then, since a and b are positive, t is positive. Let me substitute into the constraint:

a² * (ta) * (2a + ta) = a² * ta * a(2 + t) = a^4 t (2 + t) = 1.

So, a^4 t (2 + t) = 1. Therefore, a^4 = 1 / [t(2 + t)].

Then, the expression to minimize is (a + b)^2 = (a + ta)^2 = a²(1 + t)^2.

So, substituting a² from above. Since a^4 = 1/[t(2 + t)], then a² = 1 / sqrt[t(2 + t)]? Wait, no. Wait, a^4 = 1/[t(2 + t)] implies that a² = [1 / (t(2 + t))]^{1/2} = 1 / sqrt[t(2 + t)].

Therefore, the expression (a + b)^2 = a²(1 + t)^2 = [1 / sqrt(t(2 + t))] * (1 + t)^2.

So, we need to minimize [ (1 + t)^2 ] / sqrt(t(2 + t)) for t > 0.

Let me denote f(t) = (1 + t)^2 / sqrt(t(2 + t)). Let me simplify this expression. Let me square it to make it easier? Maybe not. Alternatively, write it as:

f(t) = (1 + t)^2 / [sqrt(t) * sqrt(2 + t)].

Alternatively, write it as:

f(t) = (1 + t)^2 / [sqrt(t(2 + t))].

Let me make substitution u = t, so we can take derivative. Maybe calculus is needed here. Since this seems complicated for AM-GM. Let me try calculus.

Let me compute derivative of f(t). Let me denote f(t) = (1 + t)^2 / [sqrt(t(2 + t))]. Let me write it as f(t) = (1 + t)^2 * [t(2 + t)]^{-1/2}.

Let me take natural logarithm to make differentiation easier. Let me set ln f(t) = 2 ln(1 + t) - (1/2) ln[t(2 + t)].

Then, derivative of ln f(t) is [2/(1 + t)] - (1/2)[ (1/(t(2 + t))) * ( (2 + t) + t ) ].

Wait, let me compute derivative step by step. Let me denote:

Let me compute d/dt [ln f(t)] = d/dt [2 ln(1 + t) - (1/2) ln(t(2 + t))].

So derivative is 2*(1/(1 + t)) - (1/2)*[ derivative of ln(t(2 + t)) ].

Derivative of ln(t(2 + t)) is derivative of ln t + ln(2 + t) = 1/t + 1/(2 + t).

Therefore, derivative of ln f(t) is 2/(1 + t) - (1/2)[1/t + 1/(2 + t)].

Set derivative equal to zero for minimization.

So:

2/(1 + t) - (1/2)[1/t + 1/(2 + t)] = 0.

Multiply both sides by 2 to eliminate the fraction:

4/(1 + t) - [1/t + 1/(2 + t)] = 0.

So:

4/(1 + t) = 1/t + 1/(2 + t).

Let me compute the right-hand side:

1/t + 1/(2 + t) = [ (2 + t) + t ] / [t(2 + t)] = (2 + 2t)/[t(2 + t)].

Therefore, equation becomes:

4/(1 + t) = (2 + 2t)/[t(2 + t)].

Simplify numerator on RHS: 2(1 + t). Therefore:

4/(1 + t) = 2(1 + t)/[t(2 + t)].

Multiply both sides by (1 + t):

4 = 2(1 + t)^2 / [t(2 + t)].

Multiply both sides by t(2 + t):

4t(2 + t) = 2(1 + t)^2.

Divide both sides by 2:

2t(2 + t) = (1 + t)^2.

Expand both sides:

Left side: 4t + 2t².

Right side: 1 + 2t + t².

Bring all terms to left side:

4t + 2t² - 1 - 2t - t² = 0.

Simplify:

(2t² - t²) + (4t - 2t) - 1 = 0 => t² + 2t - 1 = 0.

Solve quadratic equation t² + 2t - 1 = 0.

Solutions are t = [-2 ± sqrt(4 + 4)] / 2 = [-2 ± sqrt(8)] / 2 = [-2 ± 2*sqrt(2)] / 2 = -1 ± sqrt(2).

Since t > 0, we take the positive root: t = -1 + sqrt(2) ≈ -1 + 1.414 ≈ 0.414. Wait, sqrt(2) is approx 1.414, so -1 + sqrt(2) ≈ 0.414, which is positive. The other root is negative, so we discard it.

Therefore, critical point at t = sqrt(2) - 1.

Now, need to check if this is a minimum. Since the function tends to infinity as t approaches 0+ and as t approaches infinity, this critical point is likely to be the minimum.

So, let me compute the value of f(t) at t = sqrt(2) - 1.

First, compute t = sqrt(2) - 1. Let me compute 1 + t = 1 + sqrt(2) - 1 = sqrt(2). Also, 2 + t = 2 + sqrt(2) - 1 = 1 + sqrt(2).

Therefore, f(t) = (1 + t)^2 / sqrt(t(2 + t)) = [ (sqrt(2))^2 ] / sqrt( [sqrt(2) - 1][1 + sqrt(2)] )

Compute numerator: (sqrt(2))² = 2.

Denominator: sqrt( [sqrt(2) - 1][sqrt(2) + 1] ). Note that [sqrt(2) - 1][sqrt(2) + 1] = (sqrt(2))² - 1² = 2 - 1 = 1. Therefore, denominator is sqrt(1) = 1.

Therefore, f(t) = 2 / 1 = 2. So the minimal value of (a + b)^2 is 2. Therefore, the minimal value of (x + y)(y + z) is 2? Wait, but hold on. Wait, we assumed that x = z. So, does this mean that the minimal value is 2? But let me check if this is indeed the global minimum or if there's a lower value when x ≠ z.

But before that, let me check my steps again. So, assuming x = z, we found that the minimal value is 2. But maybe there's a lower value when x ≠ z. How can I be sure?

Alternatively, maybe the minimal value is indeed 2, achieved when x = z and t = sqrt(2) - 1. Let me check with actual numbers.

Let me compute a and b. Recall that t = b/a = sqrt(2) - 1. So, b = a*(sqrt(2) - 1). Also, from earlier, a^4 * t*(2 + t) = 1. Let me compute t*(2 + t):

t = sqrt(2) - 1, so 2 + t = 2 + sqrt(2) - 1 = 1 + sqrt(2). Therefore, t*(2 + t) = (sqrt(2) - 1)(sqrt(2) + 1) = 2 - 1 = 1. Therefore, a^4 * 1 = 1 => a^4 = 1 => a = 1 (since a > 0). Therefore, a = 1, so b = sqrt(2) - 1. Therefore, x = z = 1, y = sqrt(2) - 1.

Let me check the original constraint: xyz(x + y + z) = 1 * (sqrt(2) - 1) * 1 * (1 + sqrt(2) - 1 + 1) = (sqrt(2) - 1) * (sqrt(2) + 1) = 2 - 1 = 1. Correct. So the constraint is satisfied.

Then, (x + y)(y + z) = (1 + sqrt(2) - 1)(sqrt(2) - 1 + 1) = (sqrt(2))(sqrt(2)) = 2. Correct. So that works. So in this case, the value is 2.

But is this the minimum? Let me try another approach without assuming x = z. Maybe using Lagrange multipliers.

Let me set up the Lagrangian. Let me denote the function to minimize as f(x, y, z) = (x + y)(y + z), and the constraint is g(x, y, z) = xyz(x + y + z) - 1 = 0.

The Lagrangian is L = (x + y)(y + z) - λ(xyz(x + y + z) - 1).

Taking partial derivatives:

∂L/∂x = (y + z) - λ[yz(x + y + z) + xyz(1)] = 0.

Wait, let me compute derivative of f with respect to x:

df/dx = derivative of (x + y)(y + z) with respect to x is (1)(y + z) + 0 = y + z.

Similarly, derivative of g with respect to x is yz(x + y + z) + xyz(1) = yz(x + y + z) + xyz = yz(x + y + z + x) = yz(2x + y + z). Wait, no:

Wait, derivative of g = xyz(x + y + z) with respect to x is:

Use product rule: derivative of xyz times (x + y + z) + xyz times derivative of (x + y + z).

So:

First term: derivative of xyz with respect to x is yz*(x + y + z) + xyz*(derivative of (x + y + z) with respect to x) ?

Wait, no. Wait, actually, the function is xyz*(x + y + z). So derivative with respect to x is:

First, derivative of xyz*(x + y + z) with respect to x:

Let me denote u = xyz and v = x + y + z. Then derivative is u’v + uv’.

Compute u’ = derivative of xyz with respect to x is yz*(x + y + z) + xyz*(derivative of (x + y + z) with respect to x)? Wait, no. Wait, u = xyz, so derivative of u with respect to x is yz*(1) + 0 + 0? Wait, no. Wait, u = x * y * z. So derivative with respect to x is yz * 1 + 0 + 0 = yz. Wait, no, that's only if y and z are constants. But in reality, y and z are variables, but when taking partial derivatives, we treat other variables as constants. Wait, in the context of Lagrange multipliers, we take partial derivatives with respect to each variable, keeping others constant. Therefore, derivative of u = xyz with respect to x is yz. Then, derivative of v = x + y + z with respect to x is 1. Therefore, derivative of g with respect to x is yz*(x + y + z) + xyz*(1) = yz(x + y + z) + xyz = yz(x + y + z + x) = yz(2x + y + z). Wait, no:

Wait, it's yz(x + y + z) + xyz. Let me factor:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No, that would be if we factor yz. Wait, no:

Wait, yz(x + y + z) + xyz = yz(x + y + z + x)? No, that's not. Let me compute:

yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x)? No, that's not. Let me compute:

Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, xyz is x * y * z. So, if I factor yz, then:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No, that would be yz(x + y + z) + xyz = yz(x + y + z) + x y z = yz(x + y + z + x) ? No, that's not. Wait, maybe I need to compute it as:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Let me compute:

Let me write it as yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me take yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, maybe I need to factor something else. Let me compute:

Let me take yz(x + y + z) + xyz = yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me think of it as yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, maybe I need to compute:

Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, perhaps I need to compute:

Let me take the first term yz(x + y + z) and the second term xyz. Let me factor yz:

yz(x + y + z) + xyz = yz(x + y + z + x) ? No. Wait, if I factor yz from both terms, but the second term is xyz, which is x * yz. Therefore, it's yz(x + y + z) + x * yz = yz [ (x + y + z) + x ] = yz(2x + y + z). Ah, yes! So derivative of g with respect to x is yz(2x + y + z). Similarly, partial derivatives with respect to y and z.

Therefore, the partial derivatives of L:

∂L/∂x = (y + z) - λ yz(2x + y + z) = 0.

Similarly, ∂L/∂y = derivative of f with respect to y is (x + y)'*(y + z) + (x + y)*(y + z)' = (1)(y + z) + (x + y)(1) = y + z + x + y = x + 2y + z.

Then, derivative of g with respect to y is:

Similarly, derivative of g with respect to y is xz(2y + x + z). Wait, let me check:

Using similar logic as before, derivative of g with respect to y is:

First term: derivative of xyz*(x + y + z) with respect to y is xz*(x + y + z) + xyz*(1) = xz(x + y + z) + xyz = xz(x + y + z + y) = xz(x + 2y + z). Wait, similar to before. So:

∂g/∂y = xz(x + 2y + z).

Therefore, ∂L/∂y = (x + 2y + z) - λ xz(x + 2y + z) = 0.

Similarly, ∂L/∂z = derivative of f with respect to z is (x + y) * derivative of (y + z) with respect to z = (x + y)(1) = x + y.

Derivative of g with respect to z is xy(x + y + 2z). Therefore:

∂L/∂z = (x + y) - λ xy(x + y + 2z) = 0.

So, we have three equations:

1) (y + z) - λ yz(2x + y + z) = 0.

2) (x + 2y + z) - λ xz(x + 2y + z) = 0.

3) (x + y) - λ xy(x + y + 2z) = 0.

And the constraint equation:

4) xyz(x + y + z) = 1.

This seems complicated, but maybe we can find relations between variables. Let me see.

Let me denote S = x + y + z. Then, note that equations 1, 2, 3 involve terms like 2x + y + z, which is S + x, similarly for others.

Let me rewrite equations:

Equation 1: (y + z) = λ yz(S + x).

Equation 2: (x + 2y + z) = λ xz(S + y).

Equation 3: (x + y) = λ xy(S + z).

Note that S = x + y + z, so S + x = 2x + y + z, S + y = x + 2y + z, S + z = x + y + 2z.

Let me denote:

Equation 1: (S - x) = λ yz(S + x).

Equation 2: (S - z) = λ xz(S + y).

Equation 3: (S - y) = λ xy(S + z).

Wait, because S - x = y + z, S - z = x + y, S - y = x + z. Wait, no:

Wait, S = x + y + z. Therefore:

S - x = y + z,

S - y = x + z,

S - z = x + y.

Therefore, equations 1, 2, 3 can be rewritten as:

1) (S - x) = λ yz(S + x).

2) (S - z) = λ xz(S + y).

3) (S - y) = λ xy(S + z).

So, equations:

1) (S - x) / [yz(S + x)] = λ.

2) (S - z) / [xz(S + y)] = λ.

3) (S - y) / [xy(S + z)] = λ.

Therefore, all three expressions equal to λ. Therefore, we can set them equal to each other.

Set equation 1 equal to equation 2:

(S - x)/[yz(S + x)] = (S - z)/[xz(S + y)].

Cross multiplying:

(S - x) * xz(S + y) = (S - z) * yz(S + x).

Cancel z from both sides:

(S - x) * x(S + y) = (S - z) * y(S + x).

Similarly, divide both sides by x and y? Wait, let me write:

(S - x) * x(S + y) = (S - z) * y(S + x).

Let me rearrange:

[ (S - x) / (S + x) ] * x = [ (S - z) / (S + y) ] * y.

Hmm, not sure. Alternatively, maybe assume some symmetry. Earlier, we assumed x = z and found a solution. Let me check if that satisfies these equations.

Assume x = z. Let me denote x = z = a, y = b. Then S = a + b + a = 2a + b.

Equation 1: (S - x)/[yz(S + x)] = (2a + b - a)/[b * a (2a + b + a)] = (a + b)/[ab(3a + b)].

Equation 2: (S - z)/[xz(S + y)] = same as equation 1, since x = z. So equations 1 and 2 are equal. Equation 3: (S - y)/[xy(S + z)] = (2a + b - b)/[a * b (2a + b + a)] = (2a)/[ab(3a + b)].

So, equations 1 and 3:

Equation 1: (a + b)/[ab(3a + b)] = λ.

Equation 3: 2a/[ab(3a + b)] = λ.

Therefore, equate them:

(a + b)/[ab(3a + b)] = 2a/[ab(3a + b)].

Multiply both sides by ab(3a + b):

a + b = 2a => b = a.

Wait, but earlier when we assumed x = z, we had t = b/a = sqrt(2) - 1, which is not 1. So this suggests that if x = z, then equations 1 and 3 would require b = a. But in our previous case, we had b ≠ a. Therefore, there is inconsistency here. Wait, that suggests that my assumption of x = z might not satisfy the Lagrangian conditions unless b = a. But earlier, when I assumed x = z and found a solution, it didn't satisfy this? Wait, but in that case, we had x = z = 1, y = sqrt(2) - 1 ≈ 0.414, so b ≠ a. Therefore, there must be a mistake here.

Wait, let me check again. If x = z = a, y = b, then equation 1 and equation 3:

Equation 1: (S - x)/[yz(S + x)] = (2a + b - a)/[b * a (2a + b + a)] = (a + b)/[ab(3a + b)].

Equation 3: (S - y)/[xy(S + z)] = (2a + b - b)/[a * b (2a + b + a)] = (2a)/[ab(3a + b)].

So equate equation 1 and equation 3:

(a + b)/[ab(3a + b)] = 2a/[ab(3a + b)] => a + b = 2a => b = a. Therefore, unless b = a, equations 1 and 3 are not equal. Therefore, if we assume x = z, then in order to satisfy the Lagrangian conditions, we must have b = a. But in our previous case, we had b ≠ a. Therefore, there is a contradiction. Which suggests that my previous approach of assuming x = z might not lead to the true minimum? Or maybe I made a mistake in the Lagrangian?

Wait, but when I assumed x = z, I found a critical point for the reduced problem, but that critical point does not satisfy the Lagrangian conditions? That seems contradictory. Wait, but maybe because when I reduced the problem by assuming x = z, I might have lost some generality? Or perhaps there is a mistake in the Lagrangian approach?

Alternatively, maybe there are multiple critical points, and the one found by assuming x = z is not the one from the Lagrangian? Wait, but how?

Alternatively, maybe my mistake was in the Lagrangian equations. Let me check again.

Wait, when I set up the Lagrangian, I had:

∂L/∂x = (y + z) - λ yz(2x + y + z) = 0.

But earlier, when I assumed x = z and found a solution, let me check if that solution satisfies the Lagrangian equations.

Let me take x = z = 1, y = sqrt(2) - 1 ≈ 0.414. Then S = x + y + z = 1 + (sqrt(2) - 1) + 1 = 1 + sqrt(2).

Compute equation 1: (y + z) - λ yz(2x + y + z) = 0.

Compute y + z = (sqrt(2) - 1) + 1 = sqrt(2).

Compute 2x + y + z = 2*1 + (sqrt(2) - 1) + 1 = 2 + sqrt(2).

Compute yz = (sqrt(2) - 1)*1 = sqrt(2) - 1.

Therefore, equation 1: sqrt(2) - λ*(sqrt(2) - 1)*(2 + sqrt(2)) = 0.

Compute (sqrt(2) - 1)*(2 + sqrt(2)):

Multiply out: sqrt(2)*2 + sqrt(2)*sqrt(2) - 1*2 - 1*sqrt(2) = 2 sqrt(2) + 2 - 2 - sqrt(2) = (2 sqrt(2) - sqrt(2)) + (2 - 2) = sqrt(2) + 0 = sqrt(2).

Therefore, equation 1 becomes sqrt(2) - λ*sqrt(2) = 0 => sqrt(2)(1 - λ) = 0 => λ = 1.

Now check equation 3: (x + y) - λ xy(S + z) = 0.

Compute x + y = 1 + (sqrt(2) - 1) = sqrt(2).

Compute xy = 1*(sqrt(2) - 1) = sqrt(2) - 1.

Compute S + z = (1 + sqrt(2)) + 1 = 2 + sqrt(2).

Therefore, equation 3: sqrt(2) - λ*(sqrt(2) - 1)*(2 + sqrt(2)) = sqrt(2) - 1*(sqrt(2)) = sqrt(2) - sqrt(2) = 0. Which is correct. Similarly, equation 2 would be same as equation 1 because x = z. Therefore, equations 1, 2, 3 are satisfied with λ = 1. Therefore, my previous thought that there was inconsistency was wrong. The mistake was in my algebra when I tried to equate equations 1 and 3. When I assumed x = z, equations 1 and 3 are not necessarily equal unless b = a, but in reality, with the specific values, they are equal because of the specific relation between variables. Therefore, the solution found by assuming x = z is indeed a critical point of the Lagrangian. Therefore, that critical point is valid.

Therefore, the minimal value is indeed 2. However, to be thorough, I need to check if there are other critical points where x ≠ z which might give a lower value.

Alternatively, maybe the minimal value is indeed 2, achieved at x = z, and that is the global minimum. To confirm, let me try another approach.

Let me consider using AM-GM inequality on the original problem. Let me think.

We need to minimize (x + y)(y + z) given that xyz(x + y + z) = 1.

Let me denote A = x + y + z. Then, the constraint is xyzA = 1.

We need to minimize (x + y)(y + z). Let me note that (x + y)(y + z) = y(x + y + z) + xz = yA + xz.

Therefore, (x + y)(y + z) = yA + xz.

So, we need to minimize yA + xz, given that xyzA = 1.

Hmm, interesting. So, we have to minimize yA + xz with xyzA = 1.

Let me think about how to relate these terms. Maybe use AM-GM on yA and xz?

But yA and xz are two terms. Let me consider that. Let me denote term1 = yA, term2 = xz. Then, by AM-GM, term1 + term2 ≥ 2√(term1 * term2). Therefore:

yA + xz ≥ 2√(yA * xz) = 2√(xyzA) = 2√(1) = 2.

Therefore, (x + y)(y + z) ≥ 2. Equality holds when yA = xz.

Therefore, the minimal value is 2, achieved when yA = xz and xyzA = 1.

So, this shows that the minimal value is indeed 2, and equality occurs when yA = xz and xyzA = 1. Let me check if this is compatible with our previous solution.

In our previous solution, x = z = 1, y = sqrt(2) - 1, A = x + y + z = 1 + sqrt(2) - 1 + 1 = 1 + sqrt(2). Then yA = (sqrt(2) - 1)(1 + sqrt(2)) = 2 - 1 = 1. And xz = 1 * 1 = 1. Therefore, yA = xz = 1, which satisfies the equality condition. Therefore, this is indeed the case where equality holds. Therefore, the minimal value is 2.

Therefore, regardless of assuming x = z or using AM-GM, we arrive at the conclusion that the minimal value is 2. Therefore, the answer is 2.

But just to make sure, let me try another example. Suppose I take different values. Let me try x = 2, y = 1, z = ?

Compute constraint: xyz(x + y + z) = 2 * 1 * z * (2 + 1 + z) = 2z(3 + z) = 1. Solve for z:

2z(3 + z) = 1 => 2z² + 6z - 1 = 0 => z = [-6 ± sqrt(36 + 8)] / 4 = [-6 ± sqrt(44)] / 4 = [-6 ± 2*sqrt(11)] / 4. Taking positive root: z = [ -6 + 2*sqrt(11) ] / 4 ≈ [ -6 + 6.633 ] / 4 ≈ 0.633 / 4 ≈ 0.158. Then compute (x + y)(y + z) = (2 + 1)(1 + 0.158) ≈ 3 * 1.158 ≈ 3.474, which is greater than 2. So indeed, higher than 2.

Another test case: take x = y = z. Let me see. Let x = y = z = t. Then constraint: t * t * t * (3t) = 3t^4 = 1 => t^4 = 1/3 => t = (1/3)^{1/4} ≈ 0.7598. Then (x + y)(y + z) = (2t)(2t) = 4t² ≈ 4*(1/3)^{1/2} ≈ 4 / 1.732 ≈ 2.309, which is still greater than 2. Therefore, the minimal value of 2 seems to hold.

Therefore, I think the minimal value is indeed 2, achieved when yA = xz and xyzA = 1, which occurs when x = z and y = sqrt(2) - 1, etc.

Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To find the **minimum value** of the expression
$$
(x + y)(y + z)
$$
under the constraint
$$
xyz(x + y + z) = 1,
$$
where $ x, y, z $ are **positive real numbers**, we proceed with a combination of **algebraic manipulation**, **substitution**, and **inequality techniques**.

---

### Step 1: Expand the Expression

We begin by expanding the expression:
$$
(x + y)(y + z) = xy + xz + y^2 + yz.
$$
However, this form is not immediately helpful. Instead, we consider a more insightful approach.

---

### Step 2: Use Substitution and Symmetry

Let’s assume **symmetry** by setting $ x = z $. This is a reasonable assumption because the expression $ (x + y)(y + z) $ is symmetric in $ x $ and $ z $, and the constraint is also symmetric in $ x $ and $ z $.

Let $ x = z = a $, and $ y = b $. Then the constraint becomes:
$$
xyz(x + y + z) = a \cdot b \cdot a \cdot (a + b + a) = a^2 b (2a + b) = 1.
$$
The expression to minimize becomes:
$$
(x + y)(y + z) = (a + b)(b + a) = (a + b)^2.
$$

So, we now minimize:
$$
(a + b)^2 \quad \text{subject to} \quad a^2 b (2a + b) = 1.
$$

---

### Step 3: Use Substitution to Reduce Variables

Let’s define $ t = \frac{b}{a} $, so $ b = ta $. Then the constraint becomes:
$$
a^2 (ta) (2a + ta) = a^4 t (2 + t) = 1.
$$
So,
$$
a^4 = \frac{1}{t(2 + t)}.
$$
Then,
$$
a^2 = \frac{1}{\sqrt{t(2 + t)}}.
$$
Now, the expression to minimize becomes:
$$
(a + b)^2 = (a + ta)^2 = a^2 (1 + t)^2 = \frac{(1 + t)^2}{\sqrt{t(2 + t)}}.
$$

Let’s define:
$$
f(t) = \frac{(1 + t)^2}{\sqrt{t(2 + t)}}.
$$
We now minimize $ f(t) $ for $ t > 0 $.

---

### Step 4: Minimize the Function

Let’s compute the derivative of $ f(t) $ to find the critical points. However, instead of going through the full calculus, we can use a **substitution** to simplify the expression.

Let’s try $ t = \sqrt{2} - 1 $. Then:
- $ 1 + t = \sqrt{2} $,
- $ 2 + t = 1 + \sqrt{2} $,
- $ t(2 + t) = (\sqrt{2} - 1)(1 + \sqrt{2}) = 2 - 1 = 1 $.

So,
$$
f(t) = \frac{(\sqrt{2})^2}{\sqrt{1}} = \frac{2}{1} = 2.
$$

This is a **candidate for the minimum**.

---

### Step 5: Verify the Minimum

We now verify that this is indeed the **minimum** value. Consider the **AM-GM inequality**:

We can write:
$$
(x + y)(y + z) = y(x + y + z) + xz = yA + xz,
$$
where $ A = x + y + z $.

Then, by AM-GM:
$$
yA + xz \geq 2\sqrt{yA \cdot xz} = 2\sqrt{xyzA}.
$$
But from the constraint $ xyzA = 1 $, we get:
$$
(x + y)(y + z) \geq 2\sqrt{1} = 2.
$$
Equality holds when $ yA = xz $, which is satisfied in our earlier case.

---

### Step 6: Confirm the Equality Case

Let’s check if the equality condition is satisfied in our earlier solution:

Let $ x = z = 1 $, $ y = \sqrt{2} - 1 $, then:
- $ x + y + z = 1 + (\sqrt{2} - 1) + 1 = 1 + \sqrt{2} $,
- $ yA = (\sqrt{2} - 1)(1 + \sqrt{2}) = 2 - 1 = 1 $,
- $ xz = 1 \cdot 1 = 1 $,
- So $ yA = xz $, and the constraint is satisfied:
$$
xyz(x + y + z) = 1 \cdot (\sqrt{2} - 1) \cdot 1 \cdot (1 + \sqrt{2}) = 1.
$$

Thus, the minimum value is achieved, and the value is:
$$
(x + y)(y + z) = (1 + \sqrt{2} - 1)(\sqrt{2} - 1 + 1) = \sqrt{2} \cdot \sqrt{2} = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

Solution:
We can write $(x + y)(y + z)$ as $xz + y(x + y + z).$  By AM-GM,
\[xz + y(x + y + z) \ge 2 \sqrt{(xz)y(x + y + z)} = 2 \sqrt{xyz(x + y + z)} = 2.\]Equality holds when $xz = y(x + y + z) = 1$ and $xyz(x + y + z) = 1.$  For example, we can take $x = 1,$ $y = \sqrt{2} - 1,$ and $z = 1.$  Hence, the minimum value is $\boxed{2}.$


Explanation:
Okay, so I need to find the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $. The problem gives me that:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d $ are positive integers. My task is to compute $ b $.

First, I remember that continued fractions for irrational numbers are infinite and non-repeating. Since $ \sqrt[3]{2} $ is irrational, its continued fraction will be infinite. However, I need to find the first few terms of this continued fraction, specifically the value of $ b $, which is the second term in the expansion.

Let me recall how continued fractions work. A continued fraction is an expression of the form:

$$
x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dotsb}}}
$$

where each $ a_i $ is a positive integer. In this case, the problem is written as:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

So comparing with the standard continued fraction notation, $ a_0 = a $, $ a_1 = b $, $ a_2 = c $, etc. Therefore, I need to find the continued fraction expansion of $ \sqrt[3]{2} $ up to the second term, which is $ b $.

To find the continued fraction expansion of a number, the standard method is to take the integer part, subtract it, take the reciprocal, and repeat the process. Let me try applying this method step by step.

First, let me compute $ a $, which is the integer part of $ \sqrt[3]{2} $.

I know that $ 1^3 = 1 $ and $ 2^3 = 8 $, so $ \sqrt[3]{2} $ is between 1 and 2. Therefore, the integer part $ a $ is 1. So:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Now, to find $ b $, I need to compute the reciprocal of $ \sqrt[3]{2} - 1 $, which will give me the next term in the continued fraction.

Let me compute $ \sqrt[3]{2} - 1 $. Let me denote $ x = \sqrt[3]{2} $, so $ x = 1 + \frac{1}{b + \dots} $. Therefore, $ x - 1 = \frac{1}{b + \dots} $, so $ b + \dots = \frac{1}{x - 1} $.

Therefore, $ b $ is the integer part of $ \frac{1}{x - 1} $.

So let me compute $ x - 1 = \sqrt[3]{2} - 1 $. Let me approximate this value numerically to get an idea.

I know that $ \sqrt[3]{1} = 1 $, $ \sqrt[3]{8} = 2 $, so $ \sqrt[3]{2} $ is approximately 1.26... Let me check with a calculator. Wait, since I don't have a calculator here, maybe I can compute it more accurately.

Alternatively, I can use the Newton-Raphson method to approximate $ \sqrt[3]{2} $. Let me try that.

Let me denote $ f(y) = y^3 - 2 $. We want to find the root of this equation. Starting with an initial guess. Let me take $ y_0 = 1.26 $, since $ 1.26^3 = ?

Let me compute $ 1.26^3 $:

First, $ 1.26^2 = 1.5876 $, then multiply by 1.26:

1.5876 * 1.26. Let me compute that:

First, 1.5876 * 1 = 1.5876

1.5876 * 0.2 = 0.31752

1.5876 * 0.06 = 0.095256

Adding them up: 1.5876 + 0.31752 = 1.90512 + 0.095256 ≈ 2.000376

Wow, so $ 1.26^3 ≈ 2.000376 $, which is very close to 2. Therefore, $ \sqrt[3]{2} ≈ 1.26 $. Therefore, $ x - 1 ≈ 0.26 $, so $ 1/(x - 1) ≈ 1/0.26 ≈ 3.846 $. Therefore, the integer part of this is 3, so $ b = 3 $. But wait, let me check if this is accurate.

But maybe my approximation is too rough. Let me check with more precise calculation.

Alternatively, since $ 1.26^3 ≈ 2.000376 $, which is slightly larger than 2, so the actual value of $ \sqrt[3]{2} $ is slightly less than 1.26. Therefore, $ x - 1 $ is slightly less than 0.26, so $ 1/(x - 1) $ is slightly more than 1/0.26 ≈ 3.846. Therefore, the integer part is still 3. Therefore, $ b = 3 $.

But to be thorough, let me try to compute $ \sqrt[3]{2} $ more accurately. Let me use linear approximation.

Let me take $ y = 1.26 - \delta $, such that $ y^3 = 2 $. Let me expand $ (1.26 - \delta)^3 = 2 $.

We know that $ 1.26^3 = 2 + 0.000376 $. Therefore, expanding:

$ (1.26 - \delta)^3 = 1.26^3 - 3*(1.26)^2 * \delta + 3*(1.26)*\delta^2 - \delta^3 = 2 $.

So approximately, ignoring the higher order terms:

$ 2 + 0.000376 - 3*(1.5876)*\delta ≈ 2 $

Therefore:

$ -3*1.5876*\delta ≈ -0.000376 $

Therefore:

$ \delta ≈ 0.000376 / (3*1.5876) ≈ 0.000376 / 4.7628 ≈ 0.0000789 $

Therefore, $ \sqrt[3]{2} ≈ 1.26 - 0.0000789 ≈ 1.2599211 $

Therefore, $ x - 1 ≈ 0.2599211 $, so $ 1/(x - 1) ≈ 1 / 0.2599211 ≈ 3.847 $, which is still approximately 3.847. Therefore, the integer part is 3, so $ b = 3 $. Therefore, the first term after the integer part is 3.

But to be absolutely sure, maybe I need to perform the continued fraction expansion more formally, without relying on decimal approximations. Let me try that.

Let me denote $ x = \sqrt[3]{2} $. Then, as before, $ x = 1 + \frac{1}{b + \frac{1}{c + \dots}} $, so $ b = \lfloor \frac{1}{x - 1} \rfloor $. Let me compute $ \frac{1}{x - 1} $.

Let me compute $ x - 1 = \sqrt[3]{2} - 1 $. Let me denote $ y = x - 1 $, so $ y = \sqrt[3]{2} - 1 $. Then, $ y + 1 = \sqrt[3]{2} $. Let me cube both sides:

$ (y + 1)^3 = 2 $

Expanding:

$ y^3 + 3y^2 + 3y + 1 = 2 $

Therefore:

$ y^3 + 3y^2 + 3y - 1 = 0 $

So, $ y^3 + 3y^2 + 3y = 1 $

Hmm, not sure if this helps directly, but maybe I can use this equation to find a better approximation for $ y $, which is $ x - 1 $.

Alternatively, since I need to compute $ \frac{1}{y} $, which is $ \frac{1}{\sqrt[3]{2} - 1} $. Maybe rationalizing this expression?

Let me try to rationalize $ \frac{1}{\sqrt[3]{2} - 1} $. To rationalize denominators with cube roots, we can use the identity $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $. Let me set $ a = \sqrt[3]{2} $, $ b = 1 $. Then:

$ a^3 - b^3 = 2 - 1 = 1 = (a - b)(a^2 + ab + b^2) $

Therefore, $ \frac{1}{a - b} = \frac{a^2 + ab + b^2}{1} = a^2 + ab + b^2 $

Therefore, $ \frac{1}{\sqrt[3]{2} - 1} = (\sqrt[3]{2})^2 + \sqrt[3]{2} \cdot 1 + 1^2 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $

So, $ \frac{1}{y} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute this value numerically.

Earlier, I approximated $ \sqrt[3]{2} ≈ 1.259921 $, so $ \sqrt[3]{4} = (\sqrt[3]{2})^2 ≈ (1.259921)^2 ≈ 1.5874 $.

Therefore, adding them up:

$ \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 1.5874 + 1.259921 + 1 ≈ 3.8473 $. Which matches the previous approximation of approximately 3.847. Therefore, $ \frac{1}{y} ≈ 3.847 $, so the integer part is 3, so $ b = 3 $. Therefore, the continued fraction starts as $ 1 + 1/(3 + ...) $, so $ b = 3 $.

But to make sure that there isn't a mistake here, let me check if there's a way to compute the continued fraction more accurately. Let me try to compute the next term as well, just to see if the continued fraction expansion is consistent.

So, after finding $ a = 1 $, $ b = 3 $, then the next term is obtained by taking the reciprocal of the fractional part. Let me formalize this.

Let me denote:

Let $ x_0 = \sqrt[3]{2} $

Then,

$ a_0 = \lfloor x_0 \rfloor = 1 $

Then,

$ x_1 = \frac{1}{x_0 - a_0} = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 3.847 $

Then,

$ a_1 = \lfloor x_1 \rfloor = 3 $

Then,

$ x_2 = \frac{1}{x_1 - a_1} = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1 - 3} = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} - 2} $

Hmm, let me compute this value. Let me denote $ z = \sqrt[3]{2} $, so $ z^3 = 2 $, and $ \sqrt[3]{4} = z^2 $. Therefore, $ x_2 = \frac{1}{z^2 + z - 2} $.

Let me compute $ z^2 + z - 2 $. Since $ z ≈ 1.2599 $, then $ z^2 ≈ 1.5874 $, so $ z^2 + z ≈ 1.5874 + 1.2599 ≈ 2.8473 $, so $ z^2 + z - 2 ≈ 0.8473 $. Therefore, $ x_2 ≈ 1 / 0.8473 ≈ 1.180 $. Therefore, $ a_2 = \lfloor x_2 \rfloor = 1 $.

Then, $ x_3 = 1/(x_2 - a_2) ≈ 1/(1.180 - 1) = 1/0.180 ≈ 5.555... $, so $ a_3 = 5 $, and so on.

But since the problem only asks for $ b $, which is $ a_1 $, which we found to be 3, maybe this is sufficient? However, to be thorough, let me check if there's a possibility that my initial approximation was wrong. For example, if $ \frac{1}{x - 1} $ was actually slightly less than 3.847, maybe it's 3.847... which is still more than 3.8, so the integer part is 3. Therefore, $ b = 3 $.

Alternatively, maybe there's a way to compute this exactly? Let me think.

Alternatively, perhaps I can use the continued fraction expansion of $ \sqrt[3]{2} $, but I don't recall the exact terms. However, since the problem is from a competition or an exam, maybe there is a known continued fraction expansion for $ \sqrt[3]{2} $, but I don't remember it. Alternatively, maybe I can use the method of continued fractions step by step with exact expressions.

Let me try to compute $ x_1 = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, as we found earlier. Let me denote $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute $ x_1 $ more accurately.

Let me use more accurate approximations for $ \sqrt[3]{2} $. Let me use the Newton-Raphson method to compute $ \sqrt[3]{2} $ with higher precision.

Let me recall that Newton-Raphson for finding roots of $ f(y) = y^3 - 2 $.

Starting with an initial guess $ y_0 = 1.26 $, as before.

Compute $ f(y_0) = (1.26)^3 - 2 ≈ 2.000376 - 2 = 0.000376 $

Compute $ f'(y_0) = 3*(1.26)^2 ≈ 3*(1.5876) ≈ 4.7628 $

Next iteration: $ y_1 = y_0 - f(y_0)/f'(y_0) ≈ 1.26 - 0.000376 / 4.7628 ≈ 1.26 - 0.0000789 ≈ 1.2599211 $

Compute $ f(y_1) = (1.2599211)^3 - 2 $. Let me compute this:

First, compute $ 1.2599211^3 $:

Let me compute $ 1.2599211^2 $:

First, 1.2599211 * 1.2599211:

Let me approximate this:

Let me write 1.2599211 as approximately 1.259921.

Let me compute 1.259921 * 1.259921:

Let me use the formula $ (a + b)^2 = a^2 + 2ab + b^2 $, where a = 1.25, b = 0.009921.

Wait, maybe that's too complicated. Alternatively, use decimal multiplication:

1.259921 * 1.259921:

First, multiply 1.259921 * 1.259921:

Let me note that 1.259921 is approximately 1.26 - 0.000079, as before.

But maybe I can use the previous value. Since we know that $ y_1 ≈ 1.2599211 $, and we had $ y_1^3 ≈ 2 - 0.000376 + ... $. Wait, actually, in the Newton-Raphson step, we had:

$ y_1 = y_0 - f(y_0)/f'(y_0) $, which is supposed to be a better approximation. Therefore, $ y_1^3 $ should be very close to 2. Let me check:

Let me compute $ y_1^3 $:

Let me use the previous expansion. Since $ y_1 = y_0 - \delta $, where $ \delta ≈ 0.0000789 $, then:

$ y_1^3 ≈ (y_0 - \delta)^3 ≈ y_0^3 - 3 y_0^2 \delta $

We know that $ y_0^3 = 2 + 0.000376 $, so:

$ y_1^3 ≈ 2 + 0.000376 - 3*(1.5876)*0.0000789 ≈ 2 + 0.000376 - 3*1.5876*0.0000789 $

Compute 3*1.5876 ≈ 4.7628

Then 4.7628 * 0.0000789 ≈ 0.000376 (since 4.7628 * 0.0000789 ≈ 0.000376). Therefore, the correction term is approximately -0.000376, so:

$ y_1^3 ≈ 2 + 0.000376 - 0.000376 = 2 $. Therefore, $ y_1^3 ≈ 2 $, so $ y_1 ≈ \sqrt[3]{2} $ with high precision. Therefore, $ x - 1 = y ≈ 1.2599211 - 1 = 0.2599211 $, so $ 1/(x - 1) ≈ 1 / 0.2599211 ≈ 3.847 $, as before.

Therefore, $ x_1 = 3.847 $, so $ a_1 = 3 $, and the fractional part is approximately 0.847. Therefore, $ b = 3 $.

But to make sure that there isn't a mistake in assuming that $ \frac{1}{\sqrt[3]{2} - 1} $ is equal to $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $, let me verify this algebraically.

Let me compute $ (\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1) $:

Let me denote $ a = \sqrt[3]{2} $, so $ a^3 = 2 $. Then:

$ (a - 1)(a^2 + a + 1) = a^3 + a^2 + a - a^2 - a - 1 = a^3 - 1 = 2 - 1 = 1 $.

Yes! Therefore, $ \frac{1}{a - 1} = a^2 + a + 1 $, which is exactly $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Therefore, this is exact. Therefore, $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is approximately 3.847, so $ b = 3 $.

Therefore, the answer is $ b = 3 $.

But just to be thorough, let me check if there's any possible mistake in the continued fraction process. For example, sometimes when dealing with continued fractions, especially with irrational numbers, there can be some subtleties. However, since we have an exact expression for $ x_1 $, and we can compute its value numerically with high precision, and since the integer part is clearly 3, I think this is solid.

Alternatively, let me check with another approach. Suppose I assume that $ b = 3 $, then check if the continued fraction expansion with $ a = 1 $, $ b = 3 $, and then compute the next term, and see if it's consistent with the value of $ \sqrt[3]{2} $.

So, if we take the continued fraction up to $ b $, then:

$ 1 + \frac{1}{3 + \frac{1}{c + \dots}} $

Let me compute the value of this continued fraction truncated at $ b $, i.e., $ 1 + \frac{1}{3} = 1.333... $. However, we know that $ \sqrt[3]{2} ≈ 1.2599 $, which is less than 1.333..., so the continued fraction must have more terms to approach the actual value. Therefore, the next term $ c $ must be such that when we take $ 1 + 1/(3 + 1/c) $, it gets closer to $ \sqrt[3]{2} $.

Let me compute $ 1 + 1/(3 + 1/c) $. Let me denote $ c $ as some integer. Let me suppose that $ c = 1 $, then $ 1 + 1/(3 + 1/1) = 1 + 1/(4) = 1.25 $. Which is less than $ \sqrt[3]{2} ≈ 1.2599 $. If $ c = 2 $, then $ 1 + 1/(3 + 1/2) = 1 + 1/(3.5) ≈ 1 + 0.2857 ≈ 1.2857 $, which is higher than 1.2599. Therefore, the actual value of $ c $ must be between 1 and 2? But $ c $ is a positive integer, so $ c = 1 $ gives 1.25, which is less than $ \sqrt[3]{2} $, and $ c = 2 $ gives 1.2857, which is higher. Therefore, the continued fraction must have $ c = 1 $, and then the next term would adjust accordingly. However, since we are only asked for $ b $, which is 3, this doesn't affect our answer.

But this shows that the continued fraction is converging towards $ \sqrt[3]{2} $, and our calculation of $ b = 3 $ is consistent with this.

Alternatively, let me check with the exact expression for $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 3.847 $. Therefore, the next term is $ a_1 = 3 $, and then the fractional part is $ x_1 - 3 ≈ 0.847 $, which is $ \sqrt[3]{4} + \sqrt[3]{2} + 1 - 3 = \sqrt[3]{4} + \sqrt[3]{2} - 2 ≈ 1.5874 + 1.2599 - 2 ≈ 0.8473 $, as before. Therefore, the reciprocal of this is $ 1 / 0.8473 ≈ 1.18 $, which would be the next term $ a_2 = 1 $, as we saw earlier. Therefore, the continued fraction is 1; 3, 1, ... and so on.

Therefore, all steps are consistent. Therefore, I can be confident that $ b = 3 $.

But just to make sure, let me check if there's any possible mistake in the initial step. For example, if someone thought that $ \sqrt[3]{2} $ is approximately 1.26, then $ \sqrt[3]{2} - 1 ≈ 0.26 $, so reciprocal is ≈ 3.846, so floor is 3. Therefore, $ b = 3 $. Therefore, this seems correct.

Alternatively, if I use more decimal places for $ \sqrt[3]{2} $, let me check with a better approximation. Let me use the Newton-Raphson result where $ y_1 ≈ 1.25992105 $, so $ x - 1 ≈ 0.25992105 $, so reciprocal is $ 1 / 0.25992105 ≈ 3.8473221 $. Therefore, the integer part is 3, so $ b = 3 $. Therefore, this is confirmed.

Therefore, I think the answer is 3.

**Final Answer**
\boxed{3}

To determine the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $, we begin by expressing the number in the form:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \dots $ are positive integers. Our goal is to find the value of $ b $, the second term in the continued fraction.

---

### Step 1: Determine the Integer Part $ a $

We know that:

$$
1^3 = 1 \quad \text{and} \quad 2^3 = 8
$$

So, $ \sqrt[3]{2} $ lies between 1 and 2. Therefore, the integer part $ a $ is:

$$
a = \lfloor \sqrt[3]{2} \rfloor = 1
$$

---

### Step 2: Compute the Reciprocal of the Fractional Part

Let:

$$
x = \sqrt[3]{2} = 1 + \frac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Then:

$$
x - 1 = \frac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

So:

$$
b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}} = \frac{1}{x - 1}
$$

Thus, the value of $ b $ is the **integer part** of $ \frac{1}{x - 1} $, i.e.,

$$
b = \left\lfloor \frac{1}{\sqrt[3]{2} - 1} \right\rfloor
$$

---

### Step 3: Rationalize the Denominator

Let $ y = \sqrt[3]{2} - 1 $. Then:

$$
\frac{1}{y} = \frac{1}{\sqrt[3]{2} - 1}
$$

To rationalize this, we use the identity:

$$
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
$$

Let $ a = \sqrt[3]{2} $, $ b = 1 $, so:

$$
a^3 - b^3 = 2 - 1 = 1 = (a - b)(a^2 + ab + b^2)
$$

Therefore:

$$
\frac{1}{a - b} = a^2 + ab + b^2 = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

So:

$$
\frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

---

### Step 4: Approximate the Value

We approximate the cube roots:

- $ \sqrt[3]{2} \approx 1.259921 $
- $ \sqrt[3]{4} = (\sqrt[3]{2})^2 \approx 1.5874 $

So:

$$
\sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 1.5874 + 1.259921 + 1 = 3.847321
$$

Thus:

$$
b = \left\lfloor 3.847321 \right\rfloor = 3
$$

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $. The problem gives me that:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d $ are positive integers. My task is to compute $ b $.

First, I remember that continued fractions for irrational numbers are infinite and non-repeating. Since $ \sqrt[3]{2} $ is irrational, its continued fraction will be infinite. However, I need to find the first few terms of this continued fraction, specifically the value of $ b $, which is the second term in the expansion.

Let me recall how continued fractions work. A continued fraction is an expression of the form:

$$
x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dotsb}}}
$$

where each $ a_i $ is a positive integer. In this case, the problem is written as:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

So comparing with the standard continued fraction notation, $ a_0 = a $, $ a_1 = b $, $ a_2 = c $, etc. Therefore, I need to find the continued fraction expansion of $ \sqrt[3]{2} $ up to the second term, which is $ b $.

To find the continued fraction expansion of a number, the standard method is to take the integer part, subtract it, take the reciprocal, and repeat the process. Let me try applying this method step by step.

First, let me compute $ a $, which is the integer part of $ \sqrt[3]{2} $.

I know that $ 1^3 = 1 $ and $ 2^3 = 8 $, so $ \sqrt[3]{2} $ is between 1 and 2. Therefore, the integer part $ a $ is 1. So:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Now, to find $ b $, I need to compute the reciprocal of $ \sqrt[3]{2} - 1 $, which will give me the next term in the continued fraction.

Let me compute $ \sqrt[3]{2} - 1 $. Let me denote $ x = \sqrt[3]{2} $, so $ x = 1 + \frac{1}{b + \dots} $. Therefore, $ x - 1 = \frac{1}{b + \dots} $, so $ b + \dots = \frac{1}{x - 1} $.

Therefore, $ b $ is the integer part of $ \frac{1}{x - 1} $.

So let me compute $ x - 1 = \sqrt[3]{2} - 1 $. Let me approximate this value numerically to get an idea.

I know that $ \sqrt[3]{1} = 1 $, $ \sqrt[3]{8} = 2 $, so $ \sqrt[3]{2} $ is approximately 1.26... Let me check with a calculator. Wait, since I don't have a calculator here, maybe I can compute it more accurately.

Alternatively, I can use the Newton-Raphson method to approximate $ \sqrt[3]{2} $. Let me try that.

Let me denote $ f(y) = y^3 - 2 $. We want to find the root of this equation. Starting with an initial guess. Let me take $ y_0 = 1.26 $, since $ 1.26^3 = ?

Let me compute $ 1.26^3 $:

First, $ 1.26^2 = 1.5876 $, then multiply by 1.26:

1.5876 * 1.26. Let me compute that:

First, 1.5876 * 1 = 1.5876

1.5876 * 0.2 = 0.31752

1.5876 * 0.06 = 0.095256

Adding them up: 1.5876 + 0.31752 = 1.90512 + 0.095256 ≈ 2.000376

Wow, so $ 1.26^3 ≈ 2.000376 $, which is very close to 2. Therefore, $ \sqrt[3]{2} ≈ 1.26 $. Therefore, $ x - 1 ≈ 0.26 $, so $ 1/(x - 1) ≈ 1/0.26 ≈ 3.846 $. Therefore, the integer part of this is 3, so $ b = 3 $. But wait, let me check if this is accurate.

But maybe my approximation is too rough. Let me check with more precise calculation.

Alternatively, since $ 1.26^3 ≈ 2.000376 $, which is slightly larger than 2, so the actual value of $ \sqrt[3]{2} $ is slightly less than 1.26. Therefore, $ x - 1 $ is slightly less than 0.26, so $ 1/(x - 1) $ is slightly more than 1/0.26 ≈ 3.846. Therefore, the integer part is still 3. Therefore, $ b = 3 $.

But to be thorough, let me try to compute $ \sqrt[3]{2} $ more accurately. Let me use linear approximation.

Let me take $ y = 1.26 - \delta $, such that $ y^3 = 2 $. Let me expand $ (1.26 - \delta)^3 = 2 $.

We know that $ 1.26^3 = 2 + 0.000376 $. Therefore, expanding:

$ (1.26 - \delta)^3 = 1.26^3 - 3*(1.26)^2 * \delta + 3*(1.26)*\delta^2 - \delta^3 = 2 $.

So approximately, ignoring the higher order terms:

$ 2 + 0.000376 - 3*(1.5876)*\delta ≈ 2 $

Therefore:

$ -3*1.5876*\delta ≈ -0.000376 $

Therefore:

$ \delta ≈ 0.000376 / (3*1.5876) ≈ 0.000376 / 4.7628 ≈ 0.0000789 $

Therefore, $ \sqrt[3]{2} ≈ 1.26 - 0.0000789 ≈ 1.2599211 $

Therefore, $ x - 1 ≈ 0.2599211 $, so $ 1/(x - 1) ≈ 1 / 0.2599211 ≈ 3.847 $, which is still approximately 3.847. Therefore, the integer part is 3, so $ b = 3 $. Therefore, the first term after the integer part is 3.

But to be absolutely sure, maybe I need to perform the continued fraction expansion more formally, without relying on decimal approximations. Let me try that.

Let me denote $ x = \sqrt[3]{2} $. Then, as before, $ x = 1 + \frac{1}{b + \frac{1}{c + \dots}} $, so $ b = \lfloor \frac{1}{x - 1} \rfloor $. Let me compute $ \frac{1}{x - 1} $.

Let me compute $ x - 1 = \sqrt[3]{2} - 1 $. Let me denote $ y = x - 1 $, so $ y = \sqrt[3]{2} - 1 $. Then, $ y + 1 = \sqrt[3]{2} $. Let me cube both sides:

$ (y + 1)^3 = 2 $

Expanding:

$ y^3 + 3y^2 + 3y + 1 = 2 $

Therefore:

$ y^3 + 3y^2 + 3y - 1 = 0 $

So, $ y^3 + 3y^2 + 3y = 1 $

Hmm, not sure if this helps directly, but maybe I can use this equation to find a better approximation for $ y $, which is $ x - 1 $.

Alternatively, since I need to compute $ \frac{1}{y} $, which is $ \frac{1}{\sqrt[3]{2} - 1} $. Maybe rationalizing this expression?

Let me try to rationalize $ \frac{1}{\sqrt[3]{2} - 1} $. To rationalize denominators with cube roots, we can use the identity $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $. Let me set $ a = \sqrt[3]{2} $, $ b = 1 $. Then:

$ a^3 - b^3 = 2 - 1 = 1 = (a - b)(a^2 + ab + b^2) $

Therefore, $ \frac{1}{a - b} = \frac{a^2 + ab + b^2}{1} = a^2 + ab + b^2 $

Therefore, $ \frac{1}{\sqrt[3]{2} - 1} = (\sqrt[3]{2})^2 + \sqrt[3]{2} \cdot 1 + 1^2 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $

So, $ \frac{1}{y} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute this value numerically.

Earlier, I approximated $ \sqrt[3]{2} ≈ 1.259921 $, so $ \sqrt[3]{4} = (\sqrt[3]{2})^2 ≈ (1.259921)^2 ≈ 1.5874 $.

Therefore, adding them up:

$ \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 1.5874 + 1.259921 + 1 ≈ 3.8473 $. Which matches the previous approximation of approximately 3.847. Therefore, $ \frac{1}{y} ≈ 3.847 $, so the integer part is 3, so $ b = 3 $. Therefore, the continued fraction starts as $ 1 + 1/(3 + ...) $, so $ b = 3 $.

But to make sure that there isn't a mistake here, let me check if there's a way to compute the continued fraction more accurately. Let me try to compute the next term as well, just to see if the continued fraction expansion is consistent.

So, after finding $ a = 1 $, $ b = 3 $, then the next term is obtained by taking the reciprocal of the fractional part. Let me formalize this.

Let me denote:

Let $ x_0 = \sqrt[3]{2} $

Then,

$ a_0 = \lfloor x_0 \rfloor = 1 $

Then,

$ x_1 = \frac{1}{x_0 - a_0} = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 3.847 $

Then,

$ a_1 = \lfloor x_1 \rfloor = 3 $

Then,

$ x_2 = \frac{1}{x_1 - a_1} = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1 - 3} = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} - 2} $

Hmm, let me compute this value. Let me denote $ z = \sqrt[3]{2} $, so $ z^3 = 2 $, and $ \sqrt[3]{4} = z^2 $. Therefore, $ x_2 = \frac{1}{z^2 + z - 2} $.

Let me compute $ z^2 + z - 2 $. Since $ z ≈ 1.2599 $, then $ z^2 ≈ 1.5874 $, so $ z^2 + z ≈ 1.5874 + 1.2599 ≈ 2.8473 $, so $ z^2 + z - 2 ≈ 0.8473 $. Therefore, $ x_2 ≈ 1 / 0.8473 ≈ 1.180 $. Therefore, $ a_2 = \lfloor x_2 \rfloor = 1 $.

Then, $ x_3 = 1/(x_2 - a_2) ≈ 1/(1.180 - 1) = 1/0.180 ≈ 5.555... $, so $ a_3 = 5 $, and so on.

But since the problem only asks for $ b $, which is $ a_1 $, which we found to be 3, maybe this is sufficient? However, to be thorough, let me check if there's a possibility that my initial approximation was wrong. For example, if $ \frac{1}{x - 1} $ was actually slightly less than 3.847, maybe it's 3.847... which is still more than 3.8, so the integer part is 3. Therefore, $ b = 3 $.

Alternatively, maybe there's a way to compute this exactly? Let me think.

Alternatively, perhaps I can use the continued fraction expansion of $ \sqrt[3]{2} $, but I don't recall the exact terms. However, since the problem is from a competition or an exam, maybe there is a known continued fraction expansion for $ \sqrt[3]{2} $, but I don't remember it. Alternatively, maybe I can use the method of continued fractions step by step with exact expressions.

Let me try to compute $ x_1 = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, as we found earlier. Let me denote $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute $ x_1 $ more accurately.

Let me use more accurate approximations for $ \sqrt[3]{2} $. Let me use the Newton-Raphson method to compute $ \sqrt[3]{2} $ with higher precision.

Let me recall that Newton-Raphson for finding roots of $ f(y) = y^3 - 2 $.

Starting with an initial guess $ y_0 = 1.26 $, as before.

Compute $ f(y_0) = (1.26)^3 - 2 ≈ 2.000376 - 2 = 0.000376 $

Compute $ f'(y_0) = 3*(1.26)^2 ≈ 3*(1.5876) ≈ 4.7628 $

Next iteration: $ y_1 = y_0 - f(y_0)/f'(y_0) ≈ 1.26 - 0.000376 / 4.7628 ≈ 1.26 - 0.0000789 ≈ 1.2599211 $

Compute $ f(y_1) = (1.2599211)^3 - 2 $. Let me compute this:

First, compute $ 1.2599211^3 $:

Let me compute $ 1.2599211^2 $:

First, 1.2599211 * 1.2599211:

Let me approximate this:

Let me write 1.2599211 as approximately 1.259921.

Let me compute 1.259921 * 1.259921:

Let me use the formula $ (a + b)^2 = a^2 + 2ab + b^2 $, where a = 1.25, b = 0.009921.

Wait, maybe that's too complicated. Alternatively, use decimal multiplication:

1.259921 * 1.259921:

First, multiply 1.259921 * 1.259921:

Let me note that 1.259921 is approximately 1.26 - 0.000079, as before.

But maybe I can use the previous value. Since we know that $ y_1 ≈ 1.2599211 $, and we had $ y_1^3 ≈ 2 - 0.000376 + ... $. Wait, actually, in the Newton-Raphson step, we had:

$ y_1 = y_0 - f(y_0)/f'(y_0) $, which is supposed to be a better approximation. Therefore, $ y_1^3 $ should be very close to 2. Let me check:

Let me compute $ y_1^3 $:

Let me use the previous expansion. Since $ y_1 = y_0 - \delta $, where $ \delta ≈ 0.0000789 $, then:

$ y_1^3 ≈ (y_0 - \delta)^3 ≈ y_0^3 - 3 y_0^2 \delta $

We know that $ y_0^3 = 2 + 0.000376 $, so:

$ y_1^3 ≈ 2 + 0.000376 - 3*(1.5876)*0.0000789 ≈ 2 + 0.000376 - 3*1.5876*0.0000789 $

Compute 3*1.5876 ≈ 4.7628

Then 4.7628 * 0.0000789 ≈ 0.000376 (since 4.7628 * 0.0000789 ≈ 0.000376). Therefore, the correction term is approximately -0.000376, so:

$ y_1^3 ≈ 2 + 0.000376 - 0.000376 = 2 $. Therefore, $ y_1^3 ≈ 2 $, so $ y_1 ≈ \sqrt[3]{2} $ with high precision. Therefore, $ x - 1 = y ≈ 1.2599211 - 1 = 0.2599211 $, so $ 1/(x - 1) ≈ 1 / 0.2599211 ≈ 3.847 $, as before.

Therefore, $ x_1 = 3.847 $, so $ a_1 = 3 $, and the fractional part is approximately 0.847. Therefore, $ b = 3 $.

But to make sure that there isn't a mistake in assuming that $ \frac{1}{\sqrt[3]{2} - 1} $ is equal to $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $, let me verify this algebraically.

Let me compute $ (\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1) $:

Let me denote $ a = \sqrt[3]{2} $, so $ a^3 = 2 $. Then:

$ (a - 1)(a^2 + a + 1) = a^3 + a^2 + a - a^2 - a - 1 = a^3 - 1 = 2 - 1 = 1 $.

Yes! Therefore, $ \frac{1}{a - 1} = a^2 + a + 1 $, which is exactly $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Therefore, this is exact. Therefore, $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is approximately 3.847, so $ b = 3 $.

Therefore, the answer is $ b = 3 $.

But just to be thorough, let me check if there's any possible mistake in the continued fraction process. For example, sometimes when dealing with continued fractions, especially with irrational numbers, there can be some subtleties. However, since we have an exact expression for $ x_1 $, and we can compute its value numerically with high precision, and since the integer part is clearly 3, I think this is solid.

Alternatively, let me check with another approach. Suppose I assume that $ b = 3 $, then check if the continued fraction expansion with $ a = 1 $, $ b = 3 $, and then compute the next term, and see if it's consistent with the value of $ \sqrt[3]{2} $.

So, if we take the continued fraction up to $ b $, then:

$ 1 + \frac{1}{3 + \frac{1}{c + \dots}} $

Let me compute the value of this continued fraction truncated at $ b $, i.e., $ 1 + \frac{1}{3} = 1.333... $. However, we know that $ \sqrt[3]{2} ≈ 1.2599 $, which is less than 1.333..., so the continued fraction must have more terms to approach the actual value. Therefore, the next term $ c $ must be such that when we take $ 1 + 1/(3 + 1/c) $, it gets closer to $ \sqrt[3]{2} $.

Let me compute $ 1 + 1/(3 + 1/c) $. Let me denote $ c $ as some integer. Let me suppose that $ c = 1 $, then $ 1 + 1/(3 + 1/1) = 1 + 1/(4) = 1.25 $. Which is less than $ \sqrt[3]{2} ≈ 1.2599 $. If $ c = 2 $, then $ 1 + 1/(3 + 1/2) = 1 + 1/(3.5) ≈ 1 + 0.2857 ≈ 1.2857 $, which is higher than 1.2599. Therefore, the actual value of $ c $ must be between 1 and 2? But $ c $ is a positive integer, so $ c = 1 $ gives 1.25, which is less than $ \sqrt[3]{2} $, and $ c = 2 $ gives 1.2857, which is higher. Therefore, the continued fraction must have $ c = 1 $, and then the next term would adjust accordingly. However, since we are only asked for $ b $, which is 3, this doesn't affect our answer.

But this shows that the continued fraction is converging towards $ \sqrt[3]{2} $, and our calculation of $ b = 3 $ is consistent with this.

Alternatively, let me check with the exact expression for $ x_1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 ≈ 3.847 $. Therefore, the next term is $ a_1 = 3 $, and then the fractional part is $ x_1 - 3 ≈ 0.847 $, which is $ \sqrt[3]{4} + \sqrt[3]{2} + 1 - 3 = \sqrt[3]{4} + \sqrt[3]{2} - 2 ≈ 1.5874 + 1.2599 - 2 ≈ 0.8473 $, as before. Therefore, the reciprocal of this is $ 1 / 0.8473 ≈ 1.18 $, which would be the next term $ a_2 = 1 $, as we saw earlier. Therefore, the continued fraction is 1; 3, 1, ... and so on.

Therefore, all steps are consistent. Therefore, I can be confident that $ b = 3 $.

But just to make sure, let me check if there's any possible mistake in the initial step. For example, if someone thought that $ \sqrt[3]{2} $ is approximately 1.26, then $ \sqrt[3]{2} - 1 ≈ 0.26 $, so reciprocal is ≈ 3.846, so floor is 3. Therefore, $ b = 3 $. Therefore, this seems correct.

Alternatively, if I use more decimal places for $ \sqrt[3]{2} $, let me check with a better approximation. Let me use the Newton-Raphson result where $ y_1 ≈ 1.25992105 $, so $ x - 1 ≈ 0.25992105 $, so reciprocal is $ 1 / 0.25992105 ≈ 3.8473221 $. Therefore, the integer part is 3, so $ b = 3 $. Therefore, this is confirmed.

Therefore, I think the answer is 3.

**Final Answer**
\boxed{3}

To determine the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $, we begin by expressing the number in the form:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \dots $ are positive integers. Our goal is to find the value of $ b $, the second term in the continued fraction.

---

### Step 1: Determine the Integer Part $ a $

We know that:

$$
1^3 = 1 \quad \text{and} \quad 2^3 = 8
$$

So, $ \sqrt[3]{2} $ lies between 1 and 2. Therefore, the integer part $ a $ is:

$$
a = \lfloor \sqrt[3]{2} \rfloor = 1
$$

---

### Step 2: Compute the Reciprocal of the Fractional Part

Let:

$$
x = \sqrt[3]{2} = 1 + \frac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Then:

$$
x - 1 = \frac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

So:

$$
b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}} = \frac{1}{x - 1}
$$

Thus, the value of $ b $ is the **integer part** of $ \frac{1}{x - 1} $, i.e.,

$$
b = \left\lfloor \frac{1}{\sqrt[3]{2} - 1} \right\rfloor
$$

---

### Step 3: Rationalize the Denominator

Let $ y = \sqrt[3]{2} - 1 $. Then:

$$
\frac{1}{y} = \frac{1}{\sqrt[3]{2} - 1}
$$

To rationalize this, we use the identity:

$$
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
$$

Let $ a = \sqrt[3]{2} $, $ b = 1 $, so:

$$
a^3 - b^3 = 2 - 1 = 1 = (a - b)(a^2 + ab + b^2)
$$

Therefore:

$$
\frac{1}{a - b} = a^2 + ab + b^2 = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

So:

$$
\frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

---

### Step 4: Approximate the Value

We approximate the cube roots:

- $ \sqrt[3]{2} \approx 1.259921 $
- $ \sqrt[3]{4} = (\sqrt[3]{2})^2 \approx 1.5874 $

So:

$$
\sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 1.5874 + 1.259921 + 1 = 3.847321
$$

Thus:

$$
b = \left\lfloor 3.847321 \right\rfloor = 3
$$

---

### Final Answer

$$
\boxed{3}
$$

Solution:
We know that
\[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} > a,\]and
\[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} < a + 1.\]The integer $a$ that satisfies $a < \sqrt[3]{2} < a + 1$ is $a = 1.$

Then
\[\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}},\]so
\[\frac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}.\]As before, $b$ must satisfy
\[b < \frac{1}{\sqrt[3]{2} - 1} < b + 1.\]Rationalizing the denominator, we get
\[\frac{1}{\sqrt[3]{2} - 1} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{(\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1)} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{2 - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1.\]We have that
\[\sqrt[3]{4} + \sqrt[3]{2} + 1 > 1 + 1 + 1 = 3.\]Also, $1.3^3 = 2.197 > 2$ and $1.6^3 = 4.096 > 4,$ so
\[\sqrt[3]{4} + \sqrt[3]{2} + 1 < 1.3 + 1.6 + 1 = 3.9 < 4,\]so $b = \boxed{3}.$