<Question>
If $\arccos x + \arccos 2x + \arccos 3x = \pi,$ then $x$ satisfies a cubic polynomial of the form
\[ax^3 + bx^2 + cx + d = 0,\]where $a,$ $b,$ $c,$ and $d$ are integers, and $a \neq 0.$  Find the smallest possible value of $|a| + |b| + |c| + |d|.$
</Question>

<think>
<Plan_1>
Okay, so I need to solve the equation arccos(x) + arccos(2x) + arccos(3x) = π, and then find the cubic polynomial that x satisfies, right? The problem mentions that the polynomial is of the form ax³ + bx² + cx + d = 0 with integer coefficients, and I need to find the smallest possible value of |a| + |b| + |c| + |d|. Hmm, let's start by understanding the problem step by step.

First, arccos is the inverse cosine function, which has a range of [0, π]. So each term arccos(x), arccos(2x), and arccos(3x) must be between 0 and π. But also, the arguments of arccos must be within [-1, 1]. Therefore, x, 2x, and 3x must all be in [-1, 1].

<prompt_1.1> Determine the domain of x considering the range and domain of the arccos function. </prompt_1.1>

So, the equation is arccos(x) + arccos(2x) + arccos(3x) = π. Let me think about how to approach this. Maybe I can use the property that arccos(a) + arccos(b) can be combined using trigonometric identities, but here we have three terms. Hmm. Alternatively, perhaps take the cosine of both sides? Since the sum of the three arccos terms is π, maybe using the cosine addition formula multiple times.

Let me recall that cos(A + B + C) = cos(A + B)cos(C) - sin(A + B)sin(C). But since A + B + C = π, then cos(π) = -1. So, cos(arccos(x) + arccos(2x) + arccos(3x)) = cos(π) = -1.

Alternatively, maybe use the identity for three angles summing to π. Let me think: If A + B + C = π, then cos(A + B + C) = -1.
But also, we can express cos(A + B + C) in terms of cosines and sines of individual angles. Let me write that out:

cos(arccos(x) + arccos(2x) + arccos(3x)) = cos(π) = -1.

Expanding the left side using the cosine addition formula:

<prompt_1.2> Expand the left side of the equation using the cosine addition formula. </prompt_1.2>
-----

But wait, maybe there is a cubic equation with smaller coefficients? Let's see. Suppose we consider that when we squared both sides or did substitutions, maybe there was a simpler relation. Let me check my steps again.

Alternatively, perhaps using another approach. Let me denote θ = arccos(x), φ = arccos(2x), ψ = arccos(3x). Then θ + φ + ψ = π. Then, ψ = π - θ - φ. Taking cosine of both sides, cos(ψ) = cos(π - θ - φ) = -cos(θ + φ).

<prompt_1.3> Explore an alternative approach to derive the cubic equation. </prompt_1.3>

But let's check if the cubic equation could factor into a product of lower-degree polynomials with integer coefficients. Let's try to factor 2x³ + 3x² - 3x - 2.

<prompt_1.4> Check if the cubic polynomial can be factored into lower-degree polynomials. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's see:

For arccos(x) to be defined, x must be in [-1, 1].

Similarly, for arccos(2x), 2x must be in [-1, 1], so x must be in [-0.5, 0.5].

For arccos(3x), 3x must be in [-1, 1], so x must be in [-1/3, 1/3].

Therefore, the most restrictive interval is x ∈ [-1/3, 1/3]. So x must be between -1/3 and 1/3, inclusive.
</execution_1.1>

<execution_1.2>
cos(arccos(x))cos(arccos(2x))cos(arccos(3x)) - cos(arccos(x))cos(arccos(2x))sin(arccos(2x)) - cos(arccos(x))sin(arccos(3x))cos(arccos(2x)) - sin(arccos(x))cos(arccos(2x))sin(arccos(3x)) + sin(arccos(x))sin(arccos(2x))sin(arccos(3x)) = -1.

Wait, that seems complicated. Let me check that expansion again. Maybe there's a better way.
</execution_1.2>

<execution_1.3>
But cos(ψ) is 3x, and cos(θ + φ) can be expanded as cosθ cosφ - sinθ sinφ. Therefore:

3x = -[cosθ cosφ - sinθ sinφ] = -cosθ cosφ + sinθ sinφ.

But cosθ = x, cosφ = 2x, so substituting:

3x = -x*2x + sinθ sinφ => 3x = -2x² + sinθ sinφ.

Therefore, sinθ sinφ = 3x + 2x².
</execution_1.3>

<execution_1.4>
Maybe using rational root theorem. Possible rational roots are ±1, ±2, ±1/2. Let's test x=1: 2(1)^3 + 3(1)^2 -3(1) -2 = 2 +3 -3 -2=0. Wait, x=1 is a root. So (x - 1) is a factor. Let's perform polynomial division or use synthetic division.

Divide 2x³ +3x² -3x -2 by (x -1). Wait, but if x=1 is a root, then (x -1) is a factor. Let's use synthetic division:

Coefficients: 2 | 3 | -3 | -2

Bring down 2. Multiply by 1: 2. Add to next coefficient: 3 +2=5. Multiply by1:5. Add to next coefficient: -3 +5=2. Multiply by1:2. Add to last coefficient: -2 +2=0. So the quotient is 2x² +5x +2. Therefore, 2x³ +3x² -3x -2 = (x -1)(2x² +5x +2). Then, factor the quadratic: 2x² +5x +2. Discriminant: 25 -16=9. Roots: [-5 ±3]/4. So x=(-5+3)/4= -0.5 and x=(-5-3)/4=-2. Therefore, factors are (x + 0.5)(2x +1). Wait, but 2x² +5x +2 = (2x +1)(x +2). Let me check: (2x +1)(x +2)=2x² +4x +x +2=2x² +5x +2. Yes. Therefore, the cubic factors as (x -1)(2x +1)(x +2). Therefore, roots are x=1, x=-1/2, x=-2.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but let me verify that. If x is in [-1/3, 1/3], then 2x is in [-2/3, 2/3], which is within [-1, 1], so arccos(2x) is defined. Similarly, 3x is in [-1, 1], so arccos(3x) is defined. So the domain of x is [-1/3, 1/3]. So any solution must lie within this interval.
-----

Based on execution_1.2:
Let me use the identity for cos(A + B + C). Alternatively, maybe take it step by step. Let me first compute cos(arccos(x) + arccos(2x)) and then add arccos(3x).

Let me denote α = arccos(x), β = arccos(2x), γ = arccos(3x). Then the equation is α + β + γ = π. So, α + β = π - γ. Taking cosine of both sides: cos(α + β) = cos(π - γ) = -cos(γ). Therefore, cos(α + β) = -cos(γ).

But cos(α + β) can be expanded as cosα cosβ - sinα sinβ. Also, cos(γ) is 3x. Therefore:

cosα cosβ - sinα sinβ = -3x.

But cosα = x, cosβ = 2x, so substituting:

<prompt_2.1> Solve the equation cos(α + β) = -cos(γ) using the trigonometric identities. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
x * 2x - sinα sinβ = -3x

So, 2x² - sinα sinβ = -3x

Therefore, rearranged: -sinα sinβ = -3x - 2x² => sinα sinβ = 3x + 2x².
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, sinα = sin(arccos(x)). Let's compute that. If α = arccos(x), then sinα = sqrt(1 - x²).

<prompt_3.1> Calculate sinα and sinβ in terms of x. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Similarly, sinβ = sqrt(1 - (2x)^2) = sqrt(1 - 4x²).

Therefore, sinα sinβ = sqrt(1 - x²) * sqrt(1 - 4x²) = sqrt{(1 - x²)(1 - 4x²)}.

But we also have that sinα sinβ = 3x + 2x². Therefore:

sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x².
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Now, let's square both sides to eliminate the square root:

(1 - x²)(1 - 4x²) = (3x + 2x²)^2.

<prompt_4.1> Square both sides of the equation to eliminate the square root and simplify. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Let's compute both sides.

Left side: (1 - x²)(1 - 4x²) = 1*(1) + 1*(-4x²) - x²*(1) - x²*(-4x²) = 1 - 4x² - x² + 4x⁴ = 1 - 5x² + 4x⁴.

Right side: (3x + 2x²)^2 = (2x² + 3x)^2 = (2x²)^2 + 2*(2x²)*(3x) + (3x)^2 = 4x⁴ + 12x³ + 9x².

So, setting left side equal to right side:

1 - 5x² + 4x⁴ = 4x⁴ + 12x³ + 9x².

Let's bring all terms to the left side:

1 - 5x² + 4x⁴ - 4x⁴ - 12x³ - 9x² = 0

Simplify term by term:

4x⁴ - 4x⁴ = 0

-5x² -9x² = -14x²

-12x³

+1

So the equation becomes:

-12x³ -14x² + 1 = 0

Multiply both sides by -1 to make the leading coefficient positive:

12x³ + 14x² - 1 = 0
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but let me check my steps again.

<prompt_5.1> Verify the steps taken to ensure the equation is correct. </prompt_5.1>
-----

But let me check if all roots are accounted for. The original equation had three arccos terms adding to π, but when we squared, we might have introduced extraneous solutions.

<prompt_5.2> Verify if the polynomial equation is correct and check for extraneous solutions. </prompt_5.2>
-----

Let me check if x=-2 is in the domain. Wait, original domain was x ∈ [-1/3, 1/3]. So x=-2 is outside, so it's extraneous.
Similarly, x=1 is outside the domain, so extraneous. So only x=-1/2 is in the domain? Wait, x=-1/2 is within [-1/3, 1/3]?

<prompt_5.3> Check if the roots are within the domain and verify the solution. </prompt_5.3>

But let me verify this solution in the original equation.

Compute arccos(-1/2) + arccos(2*(-1/2)) + arccos(3*(-1/2)).

<prompt_5.4> Verify the solution by substituting it back into the original equation. </prompt_5.4>
</Plan_5>

<Execution_5>
<execution_5.1>
Starting from sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x². Then squaring both sides gives (1 - x²)(1 - 4x²) = (3x + 2x²)^2. Then expanding both sides:

Left: 1 - 4x² - x² + 4x⁴ = 1 -5x² +4x⁴.

Right: 9x² +12x³ +4x⁴. Then moving everything to left:

1 -5x² +4x⁴ -9x² -12x³ -4x⁴ = 1 -14x² -12x³. So equation is 1 -14x² -12x³ =0, which is same as -12x³ -14x² +1=0, which becomes 12x³ +14x² -1=0.
</execution_5.1>

<execution_5.2>
Let me check.

Starting again from sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x². Then squaring both sides gives (1 - x²)(1 - 4x²) = (3x + 2x²)^2. Then expanding both sides:

Left: 1 -4x² -x² +4x⁴ = 1 -5x² +4x⁴.

Right: 9x² +12x³ +4x⁴.

Subtracting right from left: 1 -5x² +4x⁴ -9x² -12x³ -4x⁴ = 1 -14x² -12x³ =0. So 1 -14x² -12x³ =0, which is same as -12x³ -14x² +1=0, which is same as 12x³ +14x² -1=0. So that's correct.
</execution_5.2>

<execution_5.3>
Wait, -1/2 is -0.5, which is less than -1/3 (-0.333...), so x=-1/2 is actually outside the domain. Wait, hold on! There's a problem here. The polynomial 2x³ + 3x² - 3x - 2 = 0 has roots at x=1, x=-1/2, x=-2. But our domain for x is [-1/3, 1/3]. Therefore, none of these roots are in the domain. That means there is no solution? But the problem states that there is a solution. Hmm, that can't be right.
</execution_5.3>

<execution_5.4>
arccos(-1/2) is 2π/3 (since cos(2π/3) = -1/2).

arccos(2*(-1/2)) = arccos(-1) = π.

arccos(3*(-1/2)) = arccos(-3/2). Wait, but arccos is only defined for arguments in [-1, 1]. However, 3*(-1/2) = -3/2, which is less than -1. So arccos(3x) is arccos(-3/2), which is undefined. Wait, that's a problem. So x=-1/2 is not in the domain? Wait, earlier we thought the domain was x ∈ [-1/3, 1/3], because 3x must be ≥ -1, so x ≥ -1/3. But if x=-1/2, then 3x=-3/2 < -1, which is outside the domain. Therefore, x=-1/2 is not a valid solution. So that means there was an error in our process.

Wait, so even though algebraically we arrived at x=-1/2, but in reality, 3x must be ≥ -1, so x ≥ -1/3. Therefore, x=-1/2 is invalid. So that means our cubic equation has roots outside the domain, so we need to check which roots are valid. But in our case, the only root in the domain was x=-1/2, which is actually invalid. So does that mean there is no solution? But the problem states that there is a solution, so maybe I made a mistake.

Wait, let me check the steps again. When we squared both sides, we might have introduced extraneous roots. So even though x=-1/2 solves the equation after squaring, it doesn't satisfy the original equation because 3x is outside the domain. Therefore, we need to check which roots of the cubic equation lie within the domain x ∈ [-1/3, 1/3].

So let's analyze the cubic equation 2x³ + 3x² - 3x - 2 = 0. Wait, earlier when I tried to factor it, I ended up with (x -1)(2x +1)(x +2). But if the original equation's domain is x ∈ [-1/3, 1/3], then the possible roots in this interval would be x=-1/2 (which is -0.5, outside the domain), and x=1 (also outside). So none of the roots of the cubic equation lie within the domain. That suggests that there is no solution? But the problem states that x satisfies a cubic polynomial, implying that there is a solution. Therefore, there must be an error in my approach.

Wait, perhaps when we squared both sides, we introduced extraneous solutions, but also, maybe there is another way to solve the equation. Let me retrace the steps.

We started with arccos(x) + arccos(2x) + arccos(3x) = π. Let me denote α = arccos(x), β = arccos(2x), γ = arccos(3x). Then α + β + γ = π. Then, we used the identity that cos(α + β) = -cos(γ). Then, expanding cos(α + β) as cosα cosβ - sinα sinβ, which equals -cosγ. Then substituting the values, leading to the equation sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x². Then squaring both sides gives the quartic equation, which simplifies to a cubic. But perhaps there is another approach. Let me consider another identity.

Alternatively, since α + β + γ = π, then α + β = π - γ. Taking sine of both sides: sin(α + β) = sin(π - γ) = sinγ. So sin(α + β) = sinγ. Let's compute both sides.
</execution_5.4>
</Execution_5>

<Plan_6>
Based on execution_5.4:
Left side: sinα cosβ + cosα sinβ.

<prompt_6.1> Solve the equation using a different trigonometric identity. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Right side: sinγ = sin(arccos(3x)) = sqrt(1 - (3x)^2) = sqrt(1 - 9x²).

So, sinα cosβ + cosα sinβ = sqrt(1 - 9x²).

But sinα = sqrt(1 - x²), cosβ = 2x, cosα = x, sinβ = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²)*2x + x*sqrt(1 - 4x²) = sqrt(1 - 9x²).
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
So, 2x sqrt(1 - x²) + x sqrt(1 - 4x²) = sqrt(1 - 9x²).

This seems more complicated, but maybe we can square both sides here. Let me write the equation as:

<prompt_7.1> Square both sides of the equation to eliminate the square roots. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
2x sqrt(1 - x²) + x sqrt(1 - 4x²) = sqrt(1 - 9x²).

Let me denote A = 2x sqrt(1 - x²), B = x sqrt(1 - 4x²), so A + B = sqrt(1 - 9x²). Then, squaring both sides:

(A + B)^2 = 1 - 9x².

Expanding left side: A² + 2AB + B² = 1 - 9x².

Compute A², B², and AB.

First, A = 2x sqrt(1 - x²), so A² = 4x²(1 - x²) = 4x² - 4x⁴.

B = x sqrt(1 - 4x²), so B² = x²(1 - 4x²) = x² - 4x⁴.

AB = [2x sqrt(1 - x²)] [x sqrt(1 - 4x²)] = 2x² sqrt{(1 - x²)(1 - 4x²)}.

Therefore, A² + 2AB + B² = (4x² - 4x⁴) + 2*2x² sqrt{(1 - x²)(1 - 4x²)} + (x² - 4x⁴) = 1 - 9x².

Simplify left side:

4x² -4x⁴ + x² -4x⁴ + 4x² sqrt{(1 - x²)(1 - 4x²)} = 1 -9x².

Combine like terms:

(4x² + x²) + (-4x⁴ -4x⁴) + 4x² sqrt{(1 - x²)(1 - 4x²)} = 1 -9x².

So, 5x² -8x⁴ + 4x² sqrt{(1 - x²)(1 - 4x²)} = 1 -9x².

Let me move the polynomial terms to the right side:

4x² sqrt{(1 - x²)(1 - 4x²)} = 1 -9x² -5x² +8x⁴.

Simplify right side:

1 -14x² +8x⁴.

Therefore:

sqrt{(1 - x²)(1 - 4x²)} = (1 -14x² +8x⁴)/(4x²).

But from earlier, we had sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x². Therefore:

3x + 2x² = (1 -14x² +8x⁴)/(4x²).

Multiply both sides by 4x²:

4x²(3x + 2x²) = 1 -14x² +8x⁴.

Left side: 12x³ +8x⁴.

Therefore:

12x³ +8x⁴ = 1 -14x² +8x⁴.

Subtract 8x⁴ from both sides:

12x³ = 1 -14x².

Bring all terms to left side:

12x³ +14x² -1 =0.

Which is the same cubic equation as before: 12x³ +14x² -1 =0. So same as before.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Therefore, the minimal polynomial is 12x³ +14x² -1=0. But this has coefficients sum |12| + |14| + |0| + | -1| =12+14+0+1=27. But the problem asks for the smallest possible value of |a| + |b| + |c| + |d|. Maybe there is a way to simplify this equation or perhaps there's a mistake in the process. Let me check again.

Wait, perhaps when we squared both sides, we introduced extraneous roots, but also, maybe the polynomial can be simplified. Let me check if 12x³ +14x² -1 can be factored. Let's try rational roots. Possible roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12. Let's test x=1/2:

<prompt_8.1> Verify the roots of the polynomial and check for extraneous solutions. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
12*(1/2)^3 +14*(1/2)^2 -1 =12*(1/8) +14*(1/4) -1= 1.5 +3.5 -1=4. Not zero.

x= -1/2: 12*(-1/2)^3 +14*(-1/2)^2 -1=12*(-1/8)+14*(1/4)-1= -1.5 +3.5 -1=1. Not zero.

x=1/3: 12*(1/3)^3 +14*(1/3)^2 -1=12*(1/27)+14*(1/9)-1= (12/27)+(14/9)-1= (4/9)+(14/9)-1=18/9 -1=2-1=1. Not zero.

x= -1/3: 12*(-1/3)^3 +14*(-1/3)^2 -1=12*(-1/27)+14*(1/9)-1= -12/27 +14/9 -1= -4/9 +14/9 -1=10/9 -1=1/9. Not zero.

x=1/4: 12*(1/64)+14*(1/16)-1= 12/64 +14/16 -1= 3/16 +7/8 -1= 3/16 +14/16 -16/16=1/16. Not zero.

x= -1/4:12*(-1/4)^3 +14*(-1/4)^2 -1=12*(-1/64)+14*(1/16)-1= -12/64 +14/16 -1= -3/16 +7/8 -1= -3/16 +14/16 -16/16= (-3 +14 -16)/16= (-5)/16. Not zero.

x=1/6:12*(1/216)+14*(1/36)-1=12/216 +14/36 -1=1/18 +7/18 -1=8/18 -1=4/9 -1= -5/9. Not zero.

x= -1/6:12*(-1/6)^3 +14*(-1/6)^2 -1=12*(-1/216)+14*(1/36)-1= -12/216 +14/36 -1= -1/18 +7/18 -1=6/18 -1=1/3 -1= -2/3. Not zero.

x=1/12:12*(1/1728)+14*(1/144)-1≈ very small positive + negligible -1≈ -1. Not zero.

x= -1/12: similarly≈ -1. Not zero.

So, no rational roots. Therefore, the cubic is irreducible over rationals. Therefore, the minimal polynomial is indeed 12x³ +14x² -1=0. Therefore, the coefficients are a=12, b=14, c=0, d=-1. Therefore, the sum of absolute values is |12| + |14| + |0| + |-1| =12+14+0+1=27.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
But wait, the problem says "smallest possible value of |a| + |b| + |c| + |d|". So 27. But maybe there is a simpler polynomial? Let me check if perhaps the cubic can be divided by something. For example, if we factor out a common factor, but 12,14,0,1 have no common factors except 1. So the polynomial is already minimal. Therefore, the answer is 27.
But wait, let me check if there's a different cubic with smaller coefficients. For example, maybe dividing the entire equation by something, but since the equation is 12x³ +14x² -1=0, there's no common factor. Alternatively, perhaps the problem expects the polynomial to be monic? But the problem says "a cubic polynomial of the form ax³ +bx² +cx +d=0", so a can be non-zero integer, not necessarily 1. Therefore, 12x³ +14x² -1=0 is the minimal polynomial, with coefficients sum 27. But let me check if there's a possible error in the process.

Wait, but when we had the equation sqrt{(1 - x²)(1 - 4x²)} = 3x + 2x², and then squared it, leading to the cubic. However, when we checked x=-1/2, which solves the cubic, but is outside the domain. Therefore, perhaps there is another solution within the domain. Let's check if there is a solution in the domain x ∈ [-1/3, 1/3]. Let me consider trying to solve the cubic equation 12x³ +14x² -1=0 numerically within that interval.

<prompt_9.1> Use numerical methods to approximate the solution of the cubic equation within the domain. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
Let me compute f(x) =12x³ +14x² -1.

At x=0: f(0)=0 +0 -1=-1.

At x=1/3: f(1/3)=12*(1/27) +14*(1/9) -1= 12/27 +14/9 -1= 4/9 +14/9 -9/9= (4+14-9)/9=9/9=1.

So f(1/3)=1. Therefore, between x=0 and x=1/3, f(x) goes from -1 to 1. Therefore, by Intermediate Value Theorem, there is a root between 0 and 1/3. Also, at x=0, f(x)=-1, and at x=1/3, f(x)=1. Let's approximate it. Let's use Newton-Raphson method.

Take initial guess x0=1/3≈0.3333.

f(x0)=12*(1/3)^3 +14*(1/3)^2 -1=12*(1/27)+14*(1/9)-1≈0.4444 +1.5556 -1≈1. So f(x0)=1.

Compute f'(x)=36x² +28x.

At x0=1/3, f'(x0)=36*(1/9) +28*(1/3)=4 +28/3≈4 +9.333≈13.333.

Next approximation: x1 = x0 - f(x0)/f'(x0)= 1/3 -1/13.333≈0.3333 -0.075≈0.2583.

Compute f(x1)=12*(0.2583)^3 +14*(0.2583)^2 -1≈12*(0.0172) +14*(0.0667) -1≈0.2064 +0.9338 -1≈0.1402.

f'(x1)=36*(0.2583)^2 +28*(0.2583)≈36*(0.0667) +7.2324≈2.4 +7.2324≈9.6324.

x2 = x1 - f(x1)/f'(x1)≈0.2583 -0.1402/9.6324≈0.2583 -0.0146≈0.2437.

Compute f(x2)=12*(0.2437)^3 +14*(0.2437)^2 -1≈12*(0.0146) +14*(0.0594) -1≈0.1752 +0.8316 -1≈0.0068.

f'(x2)=36*(0.2437)^2 +28*(0.2437)≈36*(0.0594) +6.8236≈2.1384 +6.8236≈8.962.

x3 =x2 - f(x2)/f'(x2)≈0.2437 -0.0068/8.962≈0.2437 -0.00076≈0.243.

Compute f(x3)=12*(0.243)^3 +14*(0.243)^2 -1≈12*(0.0144) +14*(0.0590) -1≈0.1728 +0.826 -1≈0.0.

So the root is approximately x≈0.243. But wait, 0.243 is approximately 7/29≈0.241, but exact value might be a fraction. Let me check if x=1/4=0.25.

f(1/4)=12*(1/64)+14*(1/16)-1=12/64 +14/16 -1= 3/16 +7/8 -1= 3/16 +14/16 -16/16=1/16≈0.0625. So f(1/4)=≈0.0625. So the root is between 1/4 and 1/3. Wait, but according to previous calculation, at x=1/3≈0.333, f(x)=1, and at x=0.243, f(x)=0. So the root is approximately 0.243, which is about 17/70≈0.2429. Hmm, 17/70≈0.2429. Let me check 17/70:

17/70≈0.242857. Let's compute f(17/70):

12*(17/70)^3 +14*(17/70)^2 -1.

First, compute (17/70)^2=289/4900≈0.059.

(17/70)^3=4913/343000≈0.0143.

So 12*0.0143≈0.1716.

14*0.059≈0.826.

Total≈0.1716 +0.826 -1≈0.0. So approximately zero. Therefore, x≈17/70 is a root. But 17 and 70 are coprime. So 17/70 is the exact value? Let me check:

Compute 12*(17/70)^3 +14*(17/70)^2 -1.

First, (17)^3=4913, (70)^3=343000. So 12*4913/343000=58956/343000.

Similarly, (17)^2=289, (70)^2=4900. 14*289/4900=4046/4900.

Convert both fractions to have denominator 343000:

58956/343000 + (4046/4900)*(70/70)=4046*70/343000=283220/343000.

So total numerator:58956 +283220=342176.

Therefore, 342176/343000 -1= (342176 -343000)/343000= (-824)/343000≈-0.0024. So close to zero, but not exact. Therefore, x=17/70 is not exact. Therefore, the exact root is irrational. Therefore, the minimal polynomial is indeed the cubic we found, with coefficients sum 27.

But the problem asks for the smallest possible value of |a| + |b| + |c| + |d|. Since the polynomial is 12x³ +14x² -1=0, the coefficients are 12,14,0,-1. So sum of absolute values is 12+14+0+1=27. But is there a possibility that there is a lower sum? For example, maybe the polynomial can be divided by something, but since the polynomial is 12x³ +14x² -1, and the coefficients are coprime except for 1, so the minimal sum is indeed 27. Therefore, the answer is 27.

But wait, the problem says "the smallest possible value of |a| + |b| + |c| + |d|". So if the polynomial can be scaled down, but since the polynomial is already monic if we divide by 1, which is already done. Therefore, 27 is the minimal sum. But let me confirm once again. If the polynomial is 12x³ +14x² -1=0, then |12| + |14| + |0| + |-1| =12+14+0+1=27. Therefore, the answer is 27.
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
But wait, in the problem statement, the user wrote: "the smallest possible value of |a| + |b| + |c| + |d|". So 27 is the answer. But I need to check if there's a way to get a lower sum. For example, if the polynomial can be divided by a common factor, but 12,14,0,1 have no common factor except 1. So 12x³ +14x² -1 is the minimal polynomial, so the sum is 27. Therefore, the answer is 27.

But wait, let me check with the approximate root x≈0.243, which is about 17/70≈0.2429. Let's compute 12*(17/70)^3 +14*(17/70)^2 -1≈12*(0.0143) +14*(0.059) -1≈0.1716 +0.826 -1≈0.0, which is close. So the exact root is irrational, so the minimal polynomial is indeed the cubic we found. Therefore, the answer is 27.

**Final Answer**
\boxed{27}
</Plan_10>

<Execution_10>
No prompt to execute.
</Execution_10>

<Final_answer>
Therefore, the polynomial equation for $\arccos x + \arccos(2x) + \arccos(3x) = \pi$ with the minimal possible value of $|a| + |b| + |c| + |d|$ is $\boxed{12}$.
</Final_answer>
</think>