Explanation:
Okay, so I have this equation: arccos(x) + arccos(2x) + arccos(3x) = π. And I need to find a cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d| where the polynomial is ax³ + bx² + cx + d = 0. Alright, let me start by recalling some trigonometric identities involving arccos. Maybe if I can relate the sum of these arccos terms to something else?

First, I know that arccos(A) + arccos(B) can sometimes be simplified using cosine addition formulas. But here there are three terms. Maybe I can consider combining two of them first? Let me try.

Let me denote α = arccos(x), β = arccos(2x), γ = arccos(3x). Then the equation is α + β + γ = π. Maybe I can take the cosine of both sides? Let me try that.

Taking cosine of both sides: cos(α + β + γ) = cos(π) = -1.

But expanding cos(α + β + γ) might be complicated. Alternatively, since α + β + γ = π, maybe I can write α + β = π - γ. Then take cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ.

So cos(α + β) = -cos γ. Let me compute cos(α + β). Using the cosine addition formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos γ = cos(arccos(3x)) = 3x. So:

cos α cos β - sin α sin β = -3x.

But cos α is x, cos β is 2x. So substituting:

x * 2x - sin α sin β = -3x ⇒ 2x² - sin α sin β = -3x ⇒ sin α sin β = 2x² + 3x.

Hmm. So that's an equation involving sin α and sin β. Let me compute sin α and sin β. Since α = arccos(x), sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)²) = sqrt(1 - 4x²). So:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x.

This looks complicated. Let me square both sides to eliminate the square roots:

(1 - x²)(1 - 4x²) = (2x² + 3x)².

Expanding both sides:

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

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

So equation becomes:

1 - 5x² + 4x⁴ = 4x⁴ + 12x³ + 9x².

Subtracting 4x⁴ from both sides:

1 - 5x² = 12x³ + 9x².

Bring all terms to the left:

1 - 5x² -12x³ -9x² = 0 ⇒ 1 - 12x³ -14x² = 0 ⇒ -12x³ -14x² +1 = 0 ⇒ 12x³ +14x² -1 = 0.

Wait, so that would be a cubic equation? But the problem says that x satisfies a cubic polynomial. But hold on, I squared both sides of an equation, so I might have introduced extraneous solutions. But maybe this is the correct equation? Let me check.

Wait, but let me check my algebra again.

Left side after expansion: 1 - 5x² + 4x⁴. Right side: 4x⁴ + 12x³ + 9x². Then subtract 4x⁴ from both sides: 1 - 5x² = 12x³ + 9x². Then bringing all terms to the left: 1 -5x² -12x³ -9x² = 1 -12x³ -14x² = 0. So yes, that's right. Then rearranged: -12x³ -14x² +1 = 0, multiplying both sides by -1: 12x³ +14x² -1 = 0. So that's a cubic equation. So according to this, x satisfies 12x³ +14x² -1 = 0. Then the coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| is 12 + 14 + 0 + 1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|". So maybe there's a different cubic equation with smaller coefficients? Or maybe I made an error in my assumption?

Wait, but let me check if this equation is correct. Let me verify with an example. Suppose that x is such that arccos(x) + arccos(2x) + arccos(3x) = π. Let me test if the equation 12x³ +14x² -1 = 0 holds.

But how can I check? Maybe find approximate value of x? Let me try to find x numerically.

Let me suppose that x is positive. Since arccos(x) is defined for x in [-1,1]. But 3x must be in [-1,1], so x in [-1/3, 1/3]. But if x is positive, then x is between 0 and 1/3. Let me check if there's a solution in that interval.

Let me try x= 0. Let me compute the left side of the original equation: arccos(0) + arccos(0) + arccos(0) = π/2 + π/2 + π/2 = 3π/2, which is more than π. So not equal. Let me try x=1/3. Then arccos(1/3) + arccos(2/3) + arccos(1). arccos(1) is 0. So arccos(1/3) + arccos(2/3) ≈ 1.23096 + 0.84107 ≈ 2.072, which is more than π ≈ 3.1416? Wait, no, π ≈ 3.1416. Wait, 2.072 is less than π. Wait so if x=1/3, sum is ≈ 2.072, which is less than π. So maybe there is a solution between 0 and 1/3?

Wait, let me check. Let me think of x approaching 0. Then arccos(x) approaches π/2, arccos(2x) approaches π/2, arccos(3x) approaches π/2. So total approaches 3π/2 ≈ 4.712, which is larger than π. Wait earlier, when x=0, sum is 3π/2. When x=1/3, sum is arccos(1/3) + arccos(2/3) + arccos(1). arccos(1) is 0. So arccos(1/3) ≈ 1.23096, arccos(2/3) ≈ 0.84107, total sum ≈ 1.23096 + 0.84107 + 0 ≈ 2.072, which is less than π. So since the function is continuous, there must be some x between 0 and 1/3 where the sum equals π. So there is a solution. Let me try to approximate it.

Let me take x= 0.2. Then 2x=0.4, 3x=0.6. So arccos(0.2) ≈ 1.3694, arccos(0.4) ≈ 1.1593, arccos(0.6) ≈ 0.9273. Sum ≈ 1.3694 + 1.1593 + 0.9273 ≈ 3.456, which is more than π ≈ 3.1416. So sum is decreasing from 3π/2 ≈ 4.712 when x=0 to ≈3.456 at x=0.2, then to ≈2.072 at x=1/3. So the value crosses π somewhere between x=0.2 and x=1/3. Let me try x=0.25. Then 2x=0.5, 3x=0.75. arccos(0.25) ≈ 1.3181, arccos(0.5) = π/3 ≈ 1.0472, arccos(0.75) ≈ 0.7227. Sum ≈ 1.3181 + 1.0472 + 0.7227 ≈ 3.088, which is a bit more than π. So still more. Let me try x=0.27. Then 3x=0.81. arccos(0.27) ≈ 1.298 radians, arccos(0.54) ≈ 1.002 radians, arccos(0.81) ≈ 0.628 radians. Sum ≈ 1.298 + 1.002 + 0.628 ≈ 2.928, which is less than π ≈ 3.1416. So between x=0.25 and x=0.27. Let me try x=0.26. 2x=0.52, 3x=0.78. arccos(0.26) ≈ 1.310 radians, arccos(0.52) ≈ 1.000 radians, arccos(0.78) ≈ 0.675 radians. Sum ≈ 1.310 + 1.000 + 0.675 ≈ 2.985. Still less than π. Wait π ≈ 3.1416. Let me try x=0.24. 2x=0.48, 3x=0.72. arccos(0.24) ≈ 1.339 radians, arccos(0.48) ≈ 1.067 radians, arccos(0.72) ≈ 0.765 radians. Sum ≈ 1.339 + 1.067 + 0.765 ≈ 3.171, which is more than π. So between 0.24 and 0.26. Let me try x≈0.25. Wait at x=0.25 sum≈3.088, which is less than π? Wait π is ~3.1416. Wait 3.088 is less than π. Wait, but earlier at x=0.25, the sum was 3.088? Wait no, that's inconsistent with previous. Wait, no:

Wait when x=0.25, arccos(0.25)= approx 1.318 radians, arccos(0.5)= π/3 ≈ 1.047 radians, arccos(0.75)≈0.722 radians. Adding them: 1.318 + 1.047 + 0.722 ≈ 3.087 radians. Which is less than π (~3.1416). So between x=0.24 (sum≈3.171) and x=0.25 (sum≈3.087). Therefore, crossing π somewhere in there. Let me try x=0.245. Then 2x=0.49, 3x=0.735. arccos(0.245): Let me compute it. cos(1.33) ≈ cos(1.33). Let me check cos(1.33): calculator... Well, if I don't have a calculator, maybe approximate? Alternatively, maybe use the equation I derived: 12x³ +14x² -1 = 0. Let me check for x≈0.245. Compute 12*(0.245)^3 +14*(0.245)^2 -1.

First, 0.245^2 = 0.060025. 0.245^3 ≈ 0.245 * 0.060025 ≈ 0.014706. Then 12*0.014706 ≈ 0.176. 14*0.060025 ≈ 0.84035. Adding them: 0.176 + 0.84035 ≈ 1.016. Then subtract 1: ≈0.016. So approximately 0.016. So 12x³ +14x² -1 ≈ +0.016. So positive. So at x=0.245, the value is positive. Earlier, at x=0.25, let me compute 12*(0.25)^3 +14*(0.25)^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 = 1/16 ≈ 0.0625. Wait that's positive. Wait but earlier, when I calculated the sum of arccos terms at x=0.25, I had ≈3.088 which is less than π≈3.1416. But according to the equation, 12x³ +14x² -1 is positive. So does that mean that the equation is positive? Wait, the equation was derived from the original equation, so if x satisfies the original equation, then it must satisfy the cubic equation. However, if my approximate calculation suggests that the cubic equation is positive at x=0.25, but the original equation is less than π? Wait this seems contradictory. Maybe my derivation is wrong?

Wait let me check again. Let me go back through my steps.

Original equation: α + β + γ = π. Then I set α + β = π - γ. Then took cosine: cos(α + β) = -cos γ. Then expanded cos(α + β) = cos α cos β - sin α sin β. Then substituted cos α = x, cos β = 2x, so got x*2x - sin α sin β = -cos γ. Then cos γ = 3x, so 2x² - sin α sin β = -3x, which rearranged to sin α sin β = 2x² + 3x. Then sin α = sqrt(1 -x²), sin β = sqrt(1 -4x²), so squared both sides: (1 -x²)(1 -4x²) = (2x² + 3x)^2. Then expanding left side: 1 -5x² +4x⁴. Right side: 4x⁴ + 12x³ + 9x². Then equating them: 1 -5x² +4x⁴ = 4x⁴ +12x³ +9x². Subtract 4x⁴: 1 -5x² = 12x³ +9x². Then bringing all terms to left: 1 -5x² -12x³ -9x² = 1 -12x³ -14x² = 0 ⇒ 12x³ +14x² -1 =0. So that seems correct.

But if x=0.25 gives 12*(0.25)^3 +14*(0.25)^2 -1 = 12*(1/64) +14*(1/16) -1 = (12/64 + 14/16) -1 = (3/16 + 7/8) -1 = (3/16 + 14/16) -1 = 17/16 -1 = 1/16 ≈0.0625, which is positive. But if we plug x=0.25 into the original equation, the sum is ≈3.088 < π≈3.1416. So according to the equation 12x³ +14x² -1=0, x=0.25 gives a positive value, meaning that the actual solution should be where the cubic is zero, which would be less than 0.25? But according to my previous calculation, when x=0.24, the sum is ≈3.171>π. So, when x=0.24, the sum is greater than π, and when x=0.25, the sum is less than π. Therefore, the actual root is between 0.24 and 0.25. Let me check the cubic equation at x=0.24. 12x³ +14x² -1. Let me compute:

x=0.24:

x² = 0.0576, x³ ≈ 0.24*0.0576 ≈ 0.013824.

12x³ ≈ 12*0.013824 ≈ 0.165888

14x² ≈14*0.0576 = 0.8064

Total: 0.165888 + 0.8064 -1 ≈ 0.972288 -1 ≈ -0.0277. So negative. Therefore, between x=0.24 and 0.25, the cubic goes from negative to positive, so crosses zero. Therefore, the actual solution is there. Therefore, the cubic equation is correct. Therefore, the equation is 12x³ +14x² -1=0.

But according to the problem statement, the answer is supposed to have the minimal |a| + |b| + |c| + |d|. But here, the coefficients are a=12, b=14, c=0, d=-1. So sum is 12 +14 +0 +1=27. But maybe there's a different cubic equation with smaller coefficients? Maybe the equation can be simplified or multiplied by something?

Wait, but the equation is already cubic. Is there a possibility that my derivation missed some other equation? Or maybe I need to consider the domain of x? For example, when I squared both sides of the equation, maybe introduced extraneous solutions, but in this case, maybe the equation is correct?

Alternatively, perhaps there's another way to approach the problem, leading to a different cubic equation?

Let me think. Another approach: Let me consider taking two arccos terms first. For example, take arccos(x) + arccos(2x) = π - arccos(3x). Then take cosine of both sides. Wait, that's similar to what I did before. Alternatively, maybe take another combination?

Alternatively, suppose I set θ = arccos(x), φ = arccos(2x), ψ = arccos(3x). Then θ + φ + ψ = π. So θ + φ = π - ψ. Taking cosine: cos(θ + φ) = cos(π - ψ) = -cos ψ. Then as before, cosθ cosφ - sinθ sinφ = -cosψ. Which leads to same equation.

Alternatively, perhaps take another approach. Let me consider using sum of angles. Suppose I take θ + φ = π - ψ. Then, maybe express θ in terms of φ and ψ? Not sure.

Alternatively, let me consider using sin(θ + φ) = sin(π - ψ) = sin ψ. Maybe that would be helpful?

Let me try that. So sin(θ + φ) = sin ψ. Expanding sin(θ + φ) = sin θ cos φ + cos θ sin φ. And sin ψ = sin(arccos(3x)) = sqrt(1 - (3x)^2) = sqrt(1 - 9x²). So:

sin θ cos φ + cos θ sin φ = sqrt(1 - 9x²).

But θ = arccos(x), φ = arccos(2x). So sin θ = sqrt(1 - x²), cos φ = 2x, cos θ = x, sin φ = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²)*2x + x*sqrt(1 - 4x²) = sqrt(1 - 9x²).

So 2x sqrt(1 -x²) + x sqrt(1 -4x²) = sqrt(1 -9x²).

Hmm, this seems different from the previous equation. But perhaps we can square both sides here? Maybe?

Let me denote equation:

2x sqrt(1 -x²) + x sqrt(1 -4x²) = sqrt(1 -9x²).

Let me square both sides:

[2x sqrt(1 -x²) + x sqrt(1 -4x²)]² = 1 -9x².

Expanding left side:

[2x sqrt(1 -x²)]² + [x sqrt(1 -4x²)]² + 2*(2x sqrt(1 -x²))*(x sqrt(1 -4x²)).

Compute each term:

First term: 4x²(1 -x²).

Second term: x²(1 -4x²).

Third term: 2*(2x * x)*sqrt(1 -x²)*sqrt(1 -4x²) = 4x²*sqrt{(1 -x²)(1 -4x²)}.

Therefore, left side becomes:

4x²(1 -x²) + x²(1 -4x²) + 4x² sqrt{(1 -x²)(1 -4x²)}.

Compute the first two terms:

4x² -4x⁴ + x² -4x⁴ = (4x² + x²) + (-4x⁴ -4x⁴) = 5x² - 8x⁴.

Therefore, equation is:

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

Bring the 5x² -8x⁴ to the right side:

4x² sqrt{(1 -x²)(1 -4x²)} = 1 -9x² -5x² +8x⁴ = 1 -14x² +8x⁴.

So:

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

Now, square both sides again to eliminate the square root:

[4x²]^2 * (1 -x²)(1 -4x²) = [8x⁴ -14x² +1]^2.

Compute left side:

16x⁴*(1 -x²)(1 -4x²). Let me compute (1 -x²)(1 -4x²) again:

Which was earlier 1 -5x² +4x⁴. Therefore:

Left side: 16x⁴*(1 -5x² +4x⁴) = 16x⁴ -80x⁶ +64x⁸.

Right side: [8x⁴ -14x² +1]^2. Let me expand this:

Let me denote A = 8x⁴, B = -14x², C =1. Then (A + B + C)^2 = A² + B² + C² + 2AB + 2AC + 2BC.

Compute each term:

A² = 64x^8.

B² = 196x^4.

C² = 1.

2AB = 2*(8x⁴)*(-14x²) = -224x^6.

2AC = 2*(8x⁴)*1 = 16x⁴.

2BC = 2*(-14x²)*1 = -28x².

Therefore, altogether:

64x^8 + 196x^4 +1 -224x^6 +16x^4 -28x².

Combine like terms:

64x^8 -224x^6 + (196x^4 +16x^4) + (-28x²) +1 = 64x^8 -224x^6 + 212x^4 -28x² +1.

So now, equate left and right sides:

Left: 16x⁴ -80x⁶ +64x⁸.

Right: 64x^8 -224x^6 +212x^4 -28x² +1.

Bring all terms to left side:

16x⁴ -80x⁶ +64x⁸ -64x^8 +224x^6 -212x^4 +28x² -1 = 0.

Compute term by term:

64x⁸ -64x⁸ = 0.

-80x⁶ +224x⁶ = 144x⁶.

16x⁴ -212x⁴ = -196x⁴.

Then +28x² -1.

Therefore equation is:

144x⁶ -196x⁴ +28x² -1 =0.

Hmm, so this is a sixth-degree equation. But earlier, we had a cubic equation. So which one is correct? Well, since we squared twice, so introduced extraneous solutions. But the original equation must satisfy both equations? So perhaps the cubic equation is a factor of this sixth-degree equation?

Let me check if 12x³ +14x² -1 divides into this sixth-degree equation. Let me denote the sixth-degree equation as:

144x⁶ -196x⁴ +28x² -1 =0.

Let me perform polynomial division or see if 12x³ +14x² -1 is a factor.

Alternatively, let me substitute y = x². Then the sixth-degree equation becomes:

144y³ -196y² +28y -1 = 0.

Let me check if the cubic equation 12x³ +14x² -1 =0 relates. If I set y = x², then the cubic equation would be 12y^(3/2) +14y -1 = 0, which is not directly related. Alternatively, maybe there's a relation. Alternatively, let me see if the sixth-degree equation can be written in terms of the cubic.

Alternatively, maybe the sixth-degree equation can be factored. Let me try to factor 144y³ -196y² +28y -1.

Trying rational roots. Possible rational roots are ±1, ±1/2, ±1/3, ..., ±1/144.

Testing y=1: 144 -196 +28 -1 = (144 -196) + (28 -1) = (-52) +27 = -25 ≠0.

y=1/2: 144*(1/8) -196*(1/4) +28*(1/2) -1 = 18 -49 +14 -1 = (18+14) - (49+1)= 32 -50 = -18 ≠0.

y=1/3: 144*(1/27) -196*(1/9) +28*(1/3) -1 = (16/3) - (196/9) + (28/3) -1 = convert to ninths:

(48/9 -196/9 +84/9 -9/9) = (48 -196 +84 -9)/9 = (48 +84 -196 -9)/9 = (132 -205)/9 = (-73)/9 ≠0.

y=1/4: 144*(1/64) -196*(1/16) +28*(1/4) -1 = (9/4) - (49/4) +7 -1 = (9 -49)/4 +6 = (-40/4) +6 = -10 +6 = -4 ≠0.

y=1/6: 144*(1/216) -196*(1/36) +28*(1/6) -1 = (2/3) - (49/9) + (14/3) -1. Convert to ninths:

(6/9 -49/9 +42/9 -9/9) = (6 +42 -49 -9)/9 = (-10)/9 ≠0.

y=1/12: 144*(1/1728) -196*(1/144) +28*(1/12) -1 = 1/12 - 49/36 +7/3 -1. Convert to 36 denominator:

3/36 -49/36 +84/36 -36/36 = (3 -49 +84 -36)/36 = (3 +84) - (49 +36) = 87 -85 = 2/36 = 1/18 ≠0.

Not a rational root. Therefore, maybe the sixth-degree equation is irreducible? So then how does it relate to the cubic equation?

Alternatively, perhaps the sixth-degree equation can be factored into (quadratic)(quadratic)(quadratic) or something else? But this might be complicated.

Alternatively, since we have another equation from before, the cubic equation 12x³ +14x² -1 =0, maybe we can find a relationship between them? Let me check if the sixth-degree equation is a multiple of the cubic equation or something else.

Let me suppose that x satisfies 12x³ +14x² -1 =0. Then 12x³ = -14x² +1. Let me see if substituting this into the sixth-degree equation 144x⁶ -196x⁴ +28x² -1 =0 would make it zero.

First, compute x⁶: x⁶ = (x³)². From 12x³ = -14x² +1, so x³ = (-14x² +1)/12. Therefore, x⁶ = [(-14x² +1)/12]^2 = (196x⁴ -28x² +1)/144.

Similarly, compute x⁴: x⁴ = x * x³ = x * (-14x² +1)/12 = (-14x³ +x)/12. But x³ = (-14x² +1)/12, so substitute:

x⁴ = [ -14*(-14x² +1)/12 + x ] / 12 = [ (196x² -14)/12 + x ] / 12. Let me compute numerator:

(196x² -14)/12 + x = (196x² -14 +12x)/12. Therefore, x⁴ = [196x² +12x -14]/(12*12) = [196x² +12x -14]/144.

Now, substitute x⁶ and x⁴ into the sixth-degree equation:

144x⁶ -196x⁴ +28x² -1.

Compute each term:

144x⁶ = 144 * [ (196x⁴ -28x² +1)/144 ] = 196x⁴ -28x² +1.

-196x⁴ = -196x⁴.

So adding 144x⁶ -196x⁴:

(196x⁴ -28x² +1) -196x⁴ = -28x² +1.

Then add the remaining terms: +28x² -1.

Therefore total expression: (-28x² +1) +28x² -1 = 0. So yes! Therefore, if x satisfies 12x³ +14x² -1=0, then it satisfies the sixth-degree equation. Therefore, the cubic equation is a factor of the sixth-degree equation. Therefore, there are other roots, but the original equation is satisfied by a root of the cubic equation. So our original cubic equation is valid. Therefore, the cubic equation is correct.

Therefore, according to this, the minimal |a| + |b| + |c| + |d| is 12 +14 +0 +1=27. But the problem says "the smallest possible value of |a| + |b| + |c| + |d|". Maybe there's a different cubic equation with smaller coefficients?

But the problem says "x satisfies a cubic polynomial of the form...", so the equation 12x³ +14x² -1=0 is a cubic polynomial with integer coefficients. However, maybe there is a different cubic polynomial, perhaps with smaller coefficients, which is equivalent (i.e., scalar multiples) or maybe not? Wait, but if it's a scalar multiple, say k*(12x³ +14x² -1)=0, then the coefficients would be multiplied by k, which would increase |a| + |b| + |c| + |d|. For example, if k=1/2, then coefficients would be 6, 7, 0, -0.5, but since they need to be integers, that's not allowed. Therefore, the minimal coefficients are achieved by the original equation. However, maybe there's a different cubic equation that's not just scalar multiple but different? How can that be? Since the equation was derived through valid steps, and we checked that the cubic is a factor of the sixth-degree equation, but there might be other factors?

But if the original equation is correct, then the cubic equation is the minimal one. However, let me check again if I can find another equation. Wait, but how?

Alternatively, maybe there was an error when solving? Let me check again. From the original equation, we arrived at 12x³ +14x² -1=0. Suppose that there is another approach leading to a different cubic? Let me think.

Alternatively, using more trigonometric identities. Let me suppose that α + β + γ = π. Then maybe considering that α + β = π - γ, and then take sin(α + β) = sin(π - γ) = sin γ. Which is what I did earlier. Then led me to another equation, which after squaring twice gave sixth-degree equation. But since we already have the cubic equation, which is valid, perhaps that is the minimal one. So maybe the answer is 27? But let me check again the problem statement.

Wait, the problem says "the smallest possible value of |a| + |b| + |c| + |d|". Maybe there is another cubic equation with smaller coefficients? Let me think. For example, if the equation can be written as a cubic with smaller coefficients. Let me check if the cubic equation 12x³ +14x² -1=0 can be simplified or if there is a common factor. The coefficients are 12,14,-1. They don't have a common factor except 1. Therefore, this is the minimal polynomial with integer coefficients. Therefore, the answer would be 12 +14 +0 +1=27.

But I need to check again my initial steps. Maybe I made an error when squaring? Because sometimes squaring can introduce extraneous solutions, but the equation we obtained is correct? For example, if x is a solution to the original equation, then it must satisfy the cubic equation. However, the cubic equation may have other solutions which don't satisfy the original equation. However, since we derived the cubic equation through valid steps (assuming that the square roots are positive, which they are in the domain of x). Wait, we have to check the domain of x. Remember that arccos(x), arccos(2x), arccos(3x) must be defined. Therefore:

For arccos(x): x ∈ [-1,1].

For arccos(2x): 2x ∈ [-1,1] ⇒ x ∈ [-1/2, 1/2].

For arccos(3x): 3x ∈ [-1,1] ⇒ x ∈ [-1/3, 1/3].

Therefore, the domain of x is [-1/3, 1/3]. Also, since α + β + γ = π, each of these arccos terms must be between 0 and π, so their sum is π. Therefore, each term is between 0 and π, so x must be positive? Because arccos(x) is between 0 and π, but if x is negative, then arccos(x) is between π/2 and π. However, if x is negative, then 2x and 3x would be more negative, but arccos(2x) and arccos(3x) would require that 2x and 3x are in [-1,1]. However, if x is negative, then 2x and 3x would be negative, so arccos(2x) and arccos(3x) would be in (π/2, π). Therefore, sum of three angles each in (π/2, π) would be larger than 3π/2, which is more than π, but the original equation is sum equals π. Therefore, x must be positive. Therefore, x ∈ (0, 1/3]. Therefore, when solving, we only consider positive roots. So the cubic equation 12x³ +14x² -1=0 has one positive root in (0,1/3], as we saw numerically. Therefore, it is valid.

But just to be thorough, let me check if the cubic equation 12x³ +14x² -1=0 can be factored or transformed into another form with smaller coefficients.

For example, let me write it as:

12x³ +14x² -1 =0.

Is there a way to write this with smaller coefficients? For example, if we can factor it as something like (ax + b)(cx² + dx + e) = 0 with smaller integers?

Let me try to factor it. Suppose it factors as (mx + n)(px² + qx + r) = 0. Then:

m*p =12

m*q + n*p =14

m*r + n*q =0

n*r = -1.

Since n*r = -1, possible integer solutions are n=1, r=-1 or n=-1, r=1.

Let me try n=1, r=-1. Then equations:

m*p=12

m*q + 1*p =14

m*(-1) + 1*q =0 ⇒ -m + q =0 ⇒ q =m.

So substituting q =m into second equation:

m*m + p =14 ⇒ m² + p =14.

But m*p=12. Therefore, m and p are integers such that m*p=12, and m² + p =14.

Let me solve for p =12/m. Then equation becomes m² + 12/m =14. Multiply by m:

m³ +12 =14m ⇒ m³ -14m +12=0.

Looking for integer solutions for m. Possible divisors of 12: ±1, ±2, ±3, ±4, ±6, ±12.

Testing m=1: 1 -14 +12= -1≠0.

m=2: 8 -28 +12= -8≠0.

m=3: 27 -42 +12= -3≠0.

m=4: 64 -56 +12=20≠0.

m= -1: -1 +14 +12=25≠0.

m=-2: -8 +28 +12=32≠0.

No integer solutions. Therefore, this case doesn't work.

Now try n=-1, r=1. Then equations:

m*p =12

m*q + (-1)*p =14 ⇒ m*q -p =14.

m*r + n*q = m*1 + (-1)*q = m - q =0 ⇒ m = q.

Therefore, since m=q, substitute into second equation: m*m -p =14 ⇒ m² -p =14.

But m*p=12 ⇒ p=12/m. Therefore, m² - 12/m =14 ⇒ Multiply by m: m³ -12 =14m ⇒ m³ -14m -12=0.

Looking for integer roots. Divisors of 12: ±1, ±2, ±3, ±4, ±6, ±12.

m=1: 1 -14 -12 = -25 ≠0.

m=2: 8 -28 -12 = -32 ≠0.

m=3: 27 -42 -12 = -27 ≠0.

m=4: 64 -56 -12 = -4 ≠0.

m= -2: -8 +28 -12=8≠0.

m=-3: -27 +42 -12=3≠0.

m=-1: -1 +14 -12=1≠0.

No integer solutions. Therefore, the cubic does not factor with integer coefficients in this way. Therefore, the minimal polynomial is indeed 12x³ +14x² -1. Therefore, the sum |a| + |b| + |c| + |d| is 12 +14 +0 +1=27.

But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|". But maybe there's a different approach where you get a different cubic equation with smaller coefficients? Let me think again.

Wait, another thought. Maybe I made an error in squaring steps, which could lead to another equation, but perhaps combining the equations?

Alternatively, maybe using more identities. For example, let me consider the original equation:

α + β + γ = π. Let me use the identity for three angles summing to π. Maybe express in terms of tangent?

Alternatively, think of α + β = π - γ, and then take tan of both sides? Let me try that.

tan(α + β) = tan(π - γ) = -tan γ.

Compute tan(α + β) = [tan α + tan β]/[1 - tan α tan β] = -tan γ.

But tan α = sqrt(1 -x²)/x? Wait, tan α = sin α / cos α = sqrt(1 -x²)/x. But α = arccos(x), so tan α = sqrt(1 -x²)/x, but only if x > 0, which it is. Similarly tan β = sqrt(1 -4x²)/(2x). tan γ = sqrt(1 -9x²)/(3x). Therefore:

[sqrt(1 -x²)/x + sqrt(1 -4x²)/(2x)] / [1 - (sqrt(1 -x²)/x)(sqrt(1 -4x²)/(2x))] = -sqrt(1 -9x²)/(3x).

Multiply numerator and denominator:

Numerator: [sqrt(1 -x²)/x + sqrt(1 -4x²)/(2x)] = [2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x).

Denominator: 1 - [sqrt{(1 -x²)(1 -4x²)}]/(2x²).

Therefore, equation becomes:

[2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x) divided by [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)] equals -sqrt(1 -9x²)/(3x).

This looks very complicated. Maybe cross-multiplying? Let me denote:

Let me write:

[2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x) = [ -sqrt(1 -9x²)/(3x) ] * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

Multiply both sides by 2x:

2 sqrt(1 -x²) + sqrt(1 -4x²) = [ -sqrt(1 -9x²)/3x * 2x ] * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

Wait, simplifying the right side:

- sqrt(1 -9x²)/3x * 2x = -2 sqrt(1 -9x²)/3.

Then multiplied by [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

So:

Left side: 2 sqrt(1 -x²) + sqrt(1 -4x²).

Right side: -2 sqrt(1 -9x²)/3 * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

This seems too complex to solve without squaring again. Perhaps this approach is not helpful. Therefore, returning to the previous result.

Given that we derived the cubic equation 12x³ +14x² -1=0, and checked it through substitution and factoring attempts, and given that it's the minimal polynomial, maybe the answer is 27. However, I need to check if maybe I can scale the equation down. For example, if there's a mistake in coefficients?

Wait, let me check my algebra again when I squared the first time. Original equation after squaring was:

(1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -5x² +4x⁴.

Right side: 4x^4 + 12x³ +9x².

Subtracting 4x^4 from both sides:

1 -5x² = 12x³ +9x² ⇒ 1 -5x² -9x² -12x³ = 0 ⇒ 1 -14x² -12x³ =0 ⇒ -12x³ -14x² +1=0 ⇒ 12x³ +14x² -1=0. Yes, correct. Therefore, the equation is correct. Therefore, the coefficients are correct.

But let me check for typos. For example, is the original equation arccos(x) + arccos(2x) + arccos(3x) = π? Yes. So the steps are correct.

Alternatively, perhaps the problem is expecting a different cubic? Maybe with different coefficients?

Alternatively, maybe there is an error in the sign when expanding? Let me check again:

Original equation after taking cosine of α + β = π - γ:

cos(α + β) = -cos γ.

Which is cos α cos β - sin α sin β = -cos γ.

Substituting cos α =x, cos β =2x, cos γ =3x:

x*2x - sin α sin β = -3x ⇒ 2x² - sin α sin β = -3x ⇒ sin α sin β = 2x² +3x. Correct.

Then sin α sin β = sqrt(1 -x²) * sqrt(1 -4x²) = 2x² +3x. Then squared both sides:

(1 -x²)(1 -4x²) = (2x² +3x)^2. Which is correct. Then expanding:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 + 12x³ +9x².

Subtract 4x^4: 1 -5x² = 12x³ +9x² ⇒ 1 -5x² -9x² -12x³ = 1 -14x² -12x³ =0 ⇒ -12x³ -14x² +1=0 ⇒ 12x³ +14x² -1=0. Yes, correct.

Therefore, I think this is correct. So the answer should be 27. However, let me check if there are other possible solutions or if I missed something.

Wait, but in the problem statement, it says "x satisfies a cubic polynomial...". So perhaps there is another cubic polynomial which is satisfied by x, but with smaller coefficients? For example, if I consider that my equation 12x³ +14x² -1=0 might be equivalent to another cubic with smaller coefficients via some transformation? For example, maybe substituting y = kx or something?

Alternatively, suppose someone claims that the minimal sum is 27, but maybe there is another equation?

Alternatively, let me check if the cubic equation can be written as 12x³ +14x² -1=0 or maybe multiplied by something else?

But since we proved that this is the minimal polynomial, I think that it's safe to go with 27. However, to confirm, I can check online if similar problems have been solved. Wait, but since this is a thought process simulation, I can't actually check. However, I can recall that sometimes problems like this result in equations with smaller coefficients if approached differently, but given the steps I did are correct and verified, I think 27 is the answer.

But just to make sure, let me check with the approximate root. Earlier, I estimated that the root is between 0.24 and 0.25. Let me compute the value of 12x³ +14x² -1 at x≈0.245. As I tried earlier, it was approximately 0.016. Let me try with a better approximation. Let me use Newton-Raphson method to approximate the root.

Let me denote f(x) = 12x³ +14x² -1. We want to find x where f(x)=0.

We have f(0.24)= -0.0277 and f(0.25)=0.0625. Let me take x₀=0.245. f(0.245)= approx. Let me compute more accurately.

x=0.245

x² = 0.245² = 0.060025

x³ = 0.245 * 0.060025 ≈ 0.014706

Then:

12x³ ≈12 * 0.014706 ≈ 0.17647

14x² ≈14 * 0.060025 ≈ 0.84035

Sum: 0.17647 + 0.84035 ≈ 1.01682; subtract 1: ≈0.01682. So f(x)=0.01682 at x=0.245.

f'(x)=36x² +28x. At x=0.245:

36*(0.060025) +28*(0.245) ≈ 2.1609 + 6.86 ≈ 9.0209.

Next iteration: x₁ = x₀ - f(x₀)/f'(x₀) ≈ 0.245 - 0.01682 / 9.0209 ≈ 0.245 - 0.001865 ≈ 0.243135.

Compute f(0.243135):

x ≈0.2431

x² ≈ (0.2431)^2 ≈ 0.0591

x³ ≈ 0.2431 * 0.0591 ≈ 0.01436

Then 12x³ ≈ 0.1723, 14x²≈14*0.0591≈0.8274, total ≈0.1723 +0.8274 -1≈(0.9997) -1≈-0.0003. So f(x)≈-0.0003.

So close to zero. Then f'(x) at x≈0.2431 is 36x² +28x ≈ 36*(0.0591) +28*(0.2431) ≈ 2.1276 +6.8068≈8.9344.

Next iteration: x₁ ≈0.2431 - (-0.0003)/8.9344 ≈0.2431 +0.00003≈0.2431.

Therefore, the root is approximately 0.2431. Let me check the original equation with this value. Let me compute arccos(x) + arccos(2x) + arccos(3x) where x≈0.2431.

Compute x≈0.2431, 2x≈0.4862, 3x≈0.7293.

Compute arccos(0.2431): let me approximate. Let me recall that arccos(0.25)≈1.318 radians. Since 0.2431 is slightly less than 0.25, arccos(0.2431) will be slightly more than 1.318. Let me use calculator-like estimation.

Let me recall that cos(1.33) ≈ cos(1.33). Since π/2 ≈1.5708, so 1.33 is less than π/2, so it's in the first quadrant.

Alternatively, use a Taylor series? Maybe not. Alternatively, use linear approximation around x=0.25. Let me denote f(x)=arccos(x). At x=0.25, f(x)=1.3181. The derivative f’(x)= -1/sqrt(1 -x²). At x=0.25, f’(x)= -1/sqrt(1 -0.0625)= -1/sqrt(0.9375)≈-1/0.9682≈-1.033. Therefore, for x=0.2431, which is Δx = -0.0069 from 0.25, approximate f(x) ≈ f(0.25) + f’(0.25)*Δx ≈ 1.3181 + (-1.033)*(-0.0069) ≈ 1.3181 + 0.0071 ≈1.3252 radians.

Similarly, arccos(2x)=arccos(0.4862). Let me approximate. arccos(0.5)=π/3≈1.0472. 0.4862 is less than 0.5, so arccos(0.4862) is slightly more than π/3. Let me compute f(x)=arccos(x) at x=0.4862. Let me approximate using derivative. Let me take x=0.5, f(x)=π/3≈1.0472. f’(x)= -1/sqrt(1 -0.25)= -1/sqrt(0.75)≈-1.1547. Δx= -0.0138 (since 0.5 -0.4862=0.0138). So f(x)≈1.0472 + (-1.1547)*(-0.0138)≈1.0472 + 0.0159≈1.0631 radians.

Similarly, arccos(3x)=arccos(0.7293). Let me approximate. arccos(0.7)= approx 0.7954 radians. arccos(0.75)=approx 0.7227 radians. Since 0.7293 is between 0.7 and 0.75. Let me approximate. Let me take x=0.7293. Let me linear approximate between x=0.7 and x=0.75. Let me recall that cos(0.7954)=0.7, cos(0.7227)=0.75. So we need to find arccos(0.7293). Let me denote y=arccos(0.7293). Let me take two points:

At x1=0.7, y1=0.7954.

At x2=0.75, y2=0.7227.

We can approximate y using linear interpolation. The value 0.7293 is between 0.7 and 0.75. Let me compute the fraction:

Δx=0.75 -0.7=0.05. The value 0.7293 -0.7=0.0293. Therefore, fraction =0.0293 /0.05≈0.586. So approximate y≈y1 -0.586*(y1 - y2)= 0.7954 -0.586*(0.7954 -0.7227)= 0.7954 -0.586*(0.0727)≈0.7954 -0.0426≈0.7528 radians.

Alternatively, better to use derivative. Let me take x=0.75, y=arccos(0.75)=0.7227. The derivative f’(x)= -1/sqrt(1 -x²). At x=0.75, f’(x)= -1/sqrt(1 -0.5625)= -1/sqrt(0.4375)= -1/0.6614≈-1.511. So if we take x=0.7293, which is Δx=0.7293 -0.75= -0.0207 from x=0.75. Therefore, approximate y≈f(0.75) + f’(0.75)*Δx≈0.7227 -1.511*(-0.0207)≈0.7227 +0.0312≈0.7539 radians.

So sum of the three arccos terms≈1.3252 +1.0631 +0.7539≈3.1422 radians, which is approximately π≈3.1416. Close enough, considering the approximations. Therefore, the root x≈0.2431 satisfies the original equation, and satisfies the cubic equation 12x³ +14x² -1=0. Therefore, the equation is correct.

Hence, the minimal sum is 12 +14 +0 +1=27. Therefore, the answer should be 27.

**Final Answer**
\boxed{27}

To solve the equation:

$$
\arccos x + \arccos 2x + \arccos 3x = \pi,
$$

we define:

$$
\alpha = \arccos x,\quad \beta = \arccos 2x,\quad \gamma = \arccos 3x.
$$

Hence, we have:

$$
\alpha + \beta + \gamma = \pi.
$$

---

### Step 1: Use a Trigonometric Identity

We use the identity:

$$
\alpha + \beta = \pi - \gamma.
$$

Taking the cosine of both sides gives:

$$
\cos(\alpha + \beta) = \cos(\pi - \gamma) = -\cos \gamma.
$$

Now apply the cosine addition formula:

$$
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.
$$

Using the definitions:

$$
\cos \alpha = x,\quad \cos \beta = 2x,\quad \cos \gamma = 3x.
$$

So we get:

$$
\cos(\alpha + \beta) = x \cdot 2x - \sin \alpha \cdot \sin \beta = 2x^2 - \sin \alpha \cdot \sin \beta.
$$

Thus:

$$
2x^2 - \sin \alpha \cdot \sin \beta = -3x.
$$

Which gives:

$$
\sin \alpha \cdot \sin \beta = 2x^2 + 3x.
$$

---

### Step 2: Use Pythagorean Identities

Recall:

$$
\sin \alpha = \sqrt{1 - x^2},\quad \sin \beta = \sqrt{1 - 4x^2}.
$$

Substituting into the equation:

$$
\sqrt{1 - x^2} \cdot \sqrt{1 - 4x^2} = 2x^2 + 3x.
$$

Now square both sides:

$$
(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2.
$$

Expand both sides:

- Left-hand side:

$$
(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4.
$$

- Right-hand side:

$$
(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2.
$$

Equating both:

$$
1 - 5x^2 + 4x^4 = 4x^4 + 12x^3 + 9x^2.
$$

Subtract $ 4x^4 $ from both sides:

$$
1 - 5x^2 = 12x^3 + 9x^2.
$$

Rearrange:

$$
0 = 12x^3 + 14x^2 - 1.
$$

Hence, the equation is:

$$
12x^3 + 14x^2 - 1 = 0.
$$

---

### Step 3: Analyze the Result

We are told that $ x $ satisfies a cubic polynomial of the form:

$$
ax^3 + bx^2 + cx + d = 0,
$$

with integer coefficients and $ a \ne 0 $. Here, we obtain:

$$
a = 12,\quad b = 14,\quad c = 0,\quad d = -1.
$$

Therefore, the sum of the absolute values is:

$$
|a| + |b| + |c| + |d| = 12 + 14 + 0 + 1 = 27.
$$

---

### Step 4: Confirm Minimal Polynomial

We explored whether the cubic could be simplified (e.g., factored or scaled), but it is irreducible over the integers. Attempting rational root tests and factoring into integer polynomials did not reveal smaller coefficients, which confirms that $ 12x^3 + 14x^2 - 1 $ is indeed the minimal cubic polynomial with integer coefficients.

---

### Final Answer

$$
\boxed{27}
$$

[Assistant]
Explanation:
Okay, so I have this equation: arccos(x) + arccos(2x) + arccos(3x) = π. And I need to find a cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d| where the polynomial is ax³ + bx² + cx + d = 0. Alright, let me start by recalling some trigonometric identities involving arccos. Maybe if I can relate the sum of these arccos terms to something else?

First, I know that arccos(A) + arccos(B) can sometimes be simplified using cosine addition formulas. But here there are three terms. Maybe I can consider combining two of them first? Let me try.

Let me denote α = arccos(x), β = arccos(2x), γ = arccos(3x). Then the equation is α + β + γ = π. Maybe I can take the cosine of both sides? Let me try that.

Taking cosine of both sides: cos(α + β + γ) = cos(π) = -1.

But expanding cos(α + β + γ) might be complicated. Alternatively, since α + β + γ = π, maybe I can write α + β = π - γ. Then take cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ.

So cos(α + β) = -cos γ. Let me compute cos(α + β). Using the cosine addition formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos γ = cos(arccos(3x)) = 3x. So:

cos α cos β - sin α sin β = -3x.

But cos α is x, cos β is 2x. So substituting:

x * 2x - sin α sin β = -3x ⇒ 2x² - sin α sin β = -3x ⇒ sin α sin β = 2x² + 3x.

Hmm. So that's an equation involving sin α and sin β. Let me compute sin α and sin β. Since α = arccos(x), sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)²) = sqrt(1 - 4x²). So:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x.

This looks complicated. Let me square both sides to eliminate the square roots:

(1 - x²)(1 - 4x²) = (2x² + 3x)².

Expanding both sides:

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

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

So equation becomes:

1 - 5x² + 4x⁴ = 4x⁴ + 12x³ + 9x².

Subtracting 4x⁴ from both sides:

1 - 5x² = 12x³ + 9x².

Bring all terms to the left:

1 - 5x² -12x³ -9x² = 0 ⇒ 1 - 12x³ -14x² = 0 ⇒ -12x³ -14x² +1 = 0 ⇒ 12x³ +14x² -1 = 0.

Wait, so that would be a cubic equation? But the problem says that x satisfies a cubic polynomial. But hold on, I squared both sides of an equation, so I might have introduced extraneous solutions. But maybe this is the correct equation? Let me check.

Wait, but let me check my algebra again.

Left side after expansion: 1 - 5x² + 4x⁴. Right side: 4x⁴ + 12x³ + 9x². Then subtract 4x⁴ from both sides: 1 - 5x² = 12x³ + 9x². Then bringing all terms to the left: 1 -5x² -12x³ -9x² = 1 -12x³ -14x² = 0. So yes, that's right. Then rearranged: -12x³ -14x² +1 = 0, multiplying both sides by -1: 12x³ +14x² -1 = 0. So that's a cubic equation. So according to this, x satisfies 12x³ +14x² -1 = 0. Then the coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| is 12 + 14 + 0 + 1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|". So maybe there's a different cubic equation with smaller coefficients? Or maybe I made an error in my assumption?

Wait, but let me check if this equation is correct. Let me verify with an example. Suppose that x is such that arccos(x) + arccos(2x) + arccos(3x) = π. Let me test if the equation 12x³ +14x² -1 = 0 holds.

But how can I check? Maybe find approximate value of x? Let me try to find x numerically.

Let me suppose that x is positive. Since arccos(x) is defined for x in [-1,1]. But 3x must be in [-1,1], so x in [-1/3, 1/3]. But if x is positive, then x is between 0 and 1/3. Let me check if there's a solution in that interval.

Let me try x= 0. Let me compute the left side of the original equation: arccos(0) + arccos(0) + arccos(0) = π/2 + π/2 + π/2 = 3π/2, which is more than π. So not equal. Let me try x=1/3. Then arccos(1/3) + arccos(2/3) + arccos(1). arccos(1) is 0. So arccos(1/3) + arccos(2/3) ≈ 1.23096 + 0.84107 ≈ 2.072, which is more than π ≈ 3.1416? Wait, no, π ≈ 3.1416. Wait, 2.072 is less than π. Wait so if x=1/3, sum is ≈ 2.072, which is less than π. So maybe there is a solution between 0 and 1/3?

Wait, let me check. Let me think of x approaching 0. Then arccos(x) approaches π/2, arccos(2x) approaches π/2, arccos(3x) approaches π/2. So total approaches 3π/2 ≈ 4.712, which is larger than π. Wait earlier, when x=0, sum is 3π/2. When x=1/3, sum is arccos(1/3) + arccos(2/3) + arccos(1). arccos(1) is 0. So arccos(1/3) ≈ 1.23096, arccos(2/3) ≈ 0.84107, total sum ≈ 1.23096 + 0.84107 + 0 ≈ 2.072, which is less than π. So since the function is continuous, there must be some x between 0 and 1/3 where the sum equals π. So there is a solution. Let me try to approximate it.

Let me take x= 0.2. Then 2x=0.4, 3x=0.6. So arccos(0.2) ≈ 1.3694, arccos(0.4) ≈ 1.1593, arccos(0.6) ≈ 0.9273. Sum ≈ 1.3694 + 1.1593 + 0.9273 ≈ 3.456, which is more than π ≈ 3.1416. So sum is decreasing from 3π/2 ≈ 4.712 when x=0 to ≈3.456 at x=0.2, then to ≈2.072 at x=1/3. So the value crosses π somewhere between x=0.2 and x=1/3. Let me try x=0.25. Then 2x=0.5, 3x=0.75. arccos(0.25) ≈ 1.3181, arccos(0.5) = π/3 ≈ 1.0472, arccos(0.75) ≈ 0.7227. Sum ≈ 1.3181 + 1.0472 + 0.7227 ≈ 3.088, which is a bit more than π. So still more. Let me try x=0.27. Then 3x=0.81. arccos(0.27) ≈ 1.298 radians, arccos(0.54) ≈ 1.002 radians, arccos(0.81) ≈ 0.628 radians. Sum ≈ 1.298 + 1.002 + 0.628 ≈ 2.928, which is less than π ≈ 3.1416. So between x=0.25 and x=0.27. Let me try x=0.26. 2x=0.52, 3x=0.78. arccos(0.26) ≈ 1.310 radians, arccos(0.52) ≈ 1.000 radians, arccos(0.78) ≈ 0.675 radians. Sum ≈ 1.310 + 1.000 + 0.675 ≈ 2.985. Still less than π. Wait π ≈ 3.1416. Let me try x=0.24. 2x=0.48, 3x=0.72. arccos(0.24) ≈ 1.339 radians, arccos(0.48) ≈ 1.067 radians, arccos(0.72) ≈ 0.765 radians. Sum ≈ 1.339 + 1.067 + 0.765 ≈ 3.171, which is more than π. So between 0.24 and 0.26. Let me try x≈0.25. Wait at x=0.25 sum≈3.088, which is less than π? Wait π is ~3.1416. Wait 3.088 is less than π. Wait, but earlier at x=0.25, the sum was 3.088? Wait no, that's inconsistent with previous. Wait, no:

Wait when x=0.25, arccos(0.25)= approx 1.318 radians, arccos(0.5)= π/3 ≈ 1.047 radians, arccos(0.75)≈0.722 radians. Adding them: 1.318 + 1.047 + 0.722 ≈ 3.087 radians. Which is less than π (~3.1416). So between x=0.24 (sum≈3.171) and x=0.25 (sum≈3.087). Therefore, crossing π somewhere in there. Let me try x=0.245. Then 2x=0.49, 3x=0.735. arccos(0.245): Let me compute it. cos(1.33) ≈ cos(1.33). Let me check cos(1.33): calculator... Well, if I don't have a calculator, maybe approximate? Alternatively, maybe use the equation I derived: 12x³ +14x² -1 = 0. Let me check for x≈0.245. Compute 12*(0.245)^3 +14*(0.245)^2 -1.

First, 0.245^2 = 0.060025. 0.245^3 ≈ 0.245 * 0.060025 ≈ 0.014706. Then 12*0.014706 ≈ 0.176. 14*0.060025 ≈ 0.84035. Adding them: 0.176 + 0.84035 ≈ 1.016. Then subtract 1: ≈0.016. So approximately 0.016. So 12x³ +14x² -1 ≈ +0.016. So positive. So at x=0.245, the value is positive. Earlier, at x=0.25, let me compute 12*(0.25)^3 +14*(0.25)^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 = 1/16 ≈ 0.0625. Wait that's positive. Wait but earlier, when I calculated the sum of arccos terms at x=0.25, I had ≈3.088 which is less than π≈3.1416. But according to the equation, 12x³ +14x² -1 is positive. So does that mean that the equation is positive? Wait, the equation was derived from the original equation, so if x satisfies the original equation, then it must satisfy the cubic equation. However, if my approximate calculation suggests that the cubic equation is positive at x=0.25, but the original equation is less than π? Wait this seems contradictory. Maybe my derivation is wrong?

Wait let me check again. Let me go back through my steps.

Original equation: α + β + γ = π. Then I set α + β = π - γ. Then took cosine: cos(α + β) = -cos γ. Then expanded cos(α + β) = cos α cos β - sin α sin β. Then substituted cos α = x, cos β = 2x, so got x*2x - sin α sin β = -cos γ. Then cos γ = 3x, so 2x² - sin α sin β = -3x, which rearranged to sin α sin β = 2x² + 3x. Then sin α = sqrt(1 -x²), sin β = sqrt(1 -4x²), so squared both sides: (1 -x²)(1 -4x²) = (2x² + 3x)^2. Then expanding left side: 1 -5x² +4x⁴. Right side: 4x⁴ + 12x³ + 9x². Then equating them: 1 -5x² +4x⁴ = 4x⁴ +12x³ +9x². Subtract 4x⁴: 1 -5x² = 12x³ +9x². Then bringing all terms to left: 1 -5x² -12x³ -9x² = 1 -12x³ -14x² = 0 ⇒ 12x³ +14x² -1 =0. So that seems correct.

But if x=0.25 gives 12*(0.25)^3 +14*(0.25)^2 -1 = 12*(1/64) +14*(1/16) -1 = (12/64 + 14/16) -1 = (3/16 + 7/8) -1 = (3/16 + 14/16) -1 = 17/16 -1 = 1/16 ≈0.0625, which is positive. But if we plug x=0.25 into the original equation, the sum is ≈3.088 < π≈3.1416. So according to the equation 12x³ +14x² -1=0, x=0.25 gives a positive value, meaning that the actual solution should be where the cubic is zero, which would be less than 0.25? But according to my previous calculation, when x=0.24, the sum is ≈3.171>π. So, when x=0.24, the sum is greater than π, and when x=0.25, the sum is less than π. Therefore, the actual root is between 0.24 and 0.25. Let me check the cubic equation at x=0.24. 12x³ +14x² -1. Let me compute:

x=0.24:

x² = 0.0576, x³ ≈ 0.24*0.0576 ≈ 0.013824.

12x³ ≈ 12*0.013824 ≈ 0.165888

14x² ≈14*0.0576 = 0.8064

Total: 0.165888 + 0.8064 -1 ≈ 0.972288 -1 ≈ -0.0277. So negative. Therefore, between x=0.24 and 0.25, the cubic goes from negative to positive, so crosses zero. Therefore, the actual solution is there. Therefore, the cubic equation is correct. Therefore, the equation is 12x³ +14x² -1=0.

But according to the problem statement, the answer is supposed to have the minimal |a| + |b| + |c| + |d|. But here, the coefficients are a=12, b=14, c=0, d=-1. So sum is 12 +14 +0 +1=27. But maybe there's a different cubic equation with smaller coefficients? Maybe the equation can be simplified or multiplied by something?

Wait, but the equation is already cubic. Is there a possibility that my derivation missed some other equation? Or maybe I need to consider the domain of x? For example, when I squared both sides of the equation, maybe introduced extraneous solutions, but in this case, maybe the equation is correct?

Alternatively, perhaps there's another way to approach the problem, leading to a different cubic equation?

Let me think. Another approach: Let me consider taking two arccos terms first. For example, take arccos(x) + arccos(2x) = π - arccos(3x). Then take cosine of both sides. Wait, that's similar to what I did before. Alternatively, maybe take another combination?

Alternatively, suppose I set θ = arccos(x), φ = arccos(2x), ψ = arccos(3x). Then θ + φ + ψ = π. So θ + φ = π - ψ. Taking cosine: cos(θ + φ) = cos(π - ψ) = -cos ψ. Then as before, cosθ cosφ - sinθ sinφ = -cosψ. Which leads to same equation.

Alternatively, perhaps take another approach. Let me consider using sum of angles. Suppose I take θ + φ = π - ψ. Then, maybe express θ in terms of φ and ψ? Not sure.

Alternatively, let me consider using sin(θ + φ) = sin(π - ψ) = sin ψ. Maybe that would be helpful?

Let me try that. So sin(θ + φ) = sin ψ. Expanding sin(θ + φ) = sin θ cos φ + cos θ sin φ. And sin ψ = sin(arccos(3x)) = sqrt(1 - (3x)^2) = sqrt(1 - 9x²). So:

sin θ cos φ + cos θ sin φ = sqrt(1 - 9x²).

But θ = arccos(x), φ = arccos(2x). So sin θ = sqrt(1 - x²), cos φ = 2x, cos θ = x, sin φ = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²)*2x + x*sqrt(1 - 4x²) = sqrt(1 - 9x²).

So 2x sqrt(1 -x²) + x sqrt(1 -4x²) = sqrt(1 -9x²).

Hmm, this seems different from the previous equation. But perhaps we can square both sides here? Maybe?

Let me denote equation:

2x sqrt(1 -x²) + x sqrt(1 -4x²) = sqrt(1 -9x²).

Let me square both sides:

[2x sqrt(1 -x²) + x sqrt(1 -4x²)]² = 1 -9x².

Expanding left side:

[2x sqrt(1 -x²)]² + [x sqrt(1 -4x²)]² + 2*(2x sqrt(1 -x²))*(x sqrt(1 -4x²)).

Compute each term:

First term: 4x²(1 -x²).

Second term: x²(1 -4x²).

Third term: 2*(2x * x)*sqrt(1 -x²)*sqrt(1 -4x²) = 4x²*sqrt{(1 -x²)(1 -4x²)}.

Therefore, left side becomes:

4x²(1 -x²) + x²(1 -4x²) + 4x² sqrt{(1 -x²)(1 -4x²)}.

Compute the first two terms:

4x² -4x⁴ + x² -4x⁴ = (4x² + x²) + (-4x⁴ -4x⁴) = 5x² - 8x⁴.

Therefore, equation is:

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

Bring the 5x² -8x⁴ to the right side:

4x² sqrt{(1 -x²)(1 -4x²)} = 1 -9x² -5x² +8x⁴ = 1 -14x² +8x⁴.

So:

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

Now, square both sides again to eliminate the square root:

[4x²]^2 * (1 -x²)(1 -4x²) = [8x⁴ -14x² +1]^2.

Compute left side:

16x⁴*(1 -x²)(1 -4x²). Let me compute (1 -x²)(1 -4x²) again:

Which was earlier 1 -5x² +4x⁴. Therefore:

Left side: 16x⁴*(1 -5x² +4x⁴) = 16x⁴ -80x⁶ +64x⁸.

Right side: [8x⁴ -14x² +1]^2. Let me expand this:

Let me denote A = 8x⁴, B = -14x², C =1. Then (A + B + C)^2 = A² + B² + C² + 2AB + 2AC + 2BC.

Compute each term:

A² = 64x^8.

B² = 196x^4.

C² = 1.

2AB = 2*(8x⁴)*(-14x²) = -224x^6.

2AC = 2*(8x⁴)*1 = 16x⁴.

2BC = 2*(-14x²)*1 = -28x².

Therefore, altogether:

64x^8 + 196x^4 +1 -224x^6 +16x^4 -28x².

Combine like terms:

64x^8 -224x^6 + (196x^4 +16x^4) + (-28x²) +1 = 64x^8 -224x^6 + 212x^4 -28x² +1.

So now, equate left and right sides:

Left: 16x⁴ -80x⁶ +64x⁸.

Right: 64x^8 -224x^6 +212x^4 -28x² +1.

Bring all terms to left side:

16x⁴ -80x⁶ +64x⁸ -64x^8 +224x^6 -212x^4 +28x² -1 = 0.

Compute term by term:

64x⁸ -64x⁸ = 0.

-80x⁶ +224x⁶ = 144x⁶.

16x⁴ -212x⁴ = -196x⁴.

Then +28x² -1.

Therefore equation is:

144x⁶ -196x⁴ +28x² -1 =0.

Hmm, so this is a sixth-degree equation. But earlier, we had a cubic equation. So which one is correct? Well, since we squared twice, so introduced extraneous solutions. But the original equation must satisfy both equations? So perhaps the cubic equation is a factor of this sixth-degree equation?

Let me check if 12x³ +14x² -1 divides into this sixth-degree equation. Let me denote the sixth-degree equation as:

144x⁶ -196x⁴ +28x² -1 =0.

Let me perform polynomial division or see if 12x³ +14x² -1 is a factor.

Alternatively, let me substitute y = x². Then the sixth-degree equation becomes:

144y³ -196y² +28y -1 = 0.

Let me check if the cubic equation 12x³ +14x² -1 =0 relates. If I set y = x², then the cubic equation would be 12y^(3/2) +14y -1 = 0, which is not directly related. Alternatively, maybe there's a relation. Alternatively, let me see if the sixth-degree equation can be written in terms of the cubic.

Alternatively, maybe the sixth-degree equation can be factored. Let me try to factor 144y³ -196y² +28y -1.

Trying rational roots. Possible rational roots are ±1, ±1/2, ±1/3, ..., ±1/144.

Testing y=1: 144 -196 +28 -1 = (144 -196) + (28 -1) = (-52) +27 = -25 ≠0.

y=1/2: 144*(1/8) -196*(1/4) +28*(1/2) -1 = 18 -49 +14 -1 = (18+14) - (49+1)= 32 -50 = -18 ≠0.

y=1/3: 144*(1/27) -196*(1/9) +28*(1/3) -1 = (16/3) - (196/9) + (28/3) -1 = convert to ninths:

(48/9 -196/9 +84/9 -9/9) = (48 -196 +84 -9)/9 = (48 +84 -196 -9)/9 = (132 -205)/9 = (-73)/9 ≠0.

y=1/4: 144*(1/64) -196*(1/16) +28*(1/4) -1 = (9/4) - (49/4) +7 -1 = (9 -49)/4 +6 = (-40/4) +6 = -10 +6 = -4 ≠0.

y=1/6: 144*(1/216) -196*(1/36) +28*(1/6) -1 = (2/3) - (49/9) + (14/3) -1. Convert to ninths:

(6/9 -49/9 +42/9 -9/9) = (6 +42 -49 -9)/9 = (-10)/9 ≠0.

y=1/12: 144*(1/1728) -196*(1/144) +28*(1/12) -1 = 1/12 - 49/36 +7/3 -1. Convert to 36 denominator:

3/36 -49/36 +84/36 -36/36 = (3 -49 +84 -36)/36 = (3 +84) - (49 +36) = 87 -85 = 2/36 = 1/18 ≠0.

Not a rational root. Therefore, maybe the sixth-degree equation is irreducible? So then how does it relate to the cubic equation?

Alternatively, perhaps the sixth-degree equation can be factored into (quadratic)(quadratic)(quadratic) or something else? But this might be complicated.

Alternatively, since we have another equation from before, the cubic equation 12x³ +14x² -1 =0, maybe we can find a relationship between them? Let me check if the sixth-degree equation is a multiple of the cubic equation or something else.

Let me suppose that x satisfies 12x³ +14x² -1 =0. Then 12x³ = -14x² +1. Let me see if substituting this into the sixth-degree equation 144x⁶ -196x⁴ +28x² -1 =0 would make it zero.

First, compute x⁶: x⁶ = (x³)². From 12x³ = -14x² +1, so x³ = (-14x² +1)/12. Therefore, x⁶ = [(-14x² +1)/12]^2 = (196x⁴ -28x² +1)/144.

Similarly, compute x⁴: x⁴ = x * x³ = x * (-14x² +1)/12 = (-14x³ +x)/12. But x³ = (-14x² +1)/12, so substitute:

x⁴ = [ -14*(-14x² +1)/12 + x ] / 12 = [ (196x² -14)/12 + x ] / 12. Let me compute numerator:

(196x² -14)/12 + x = (196x² -14 +12x)/12. Therefore, x⁴ = [196x² +12x -14]/(12*12) = [196x² +12x -14]/144.

Now, substitute x⁶ and x⁴ into the sixth-degree equation:

144x⁶ -196x⁴ +28x² -1.

Compute each term:

144x⁶ = 144 * [ (196x⁴ -28x² +1)/144 ] = 196x⁴ -28x² +1.

-196x⁴ = -196x⁴.

So adding 144x⁶ -196x⁴:

(196x⁴ -28x² +1) -196x⁴ = -28x² +1.

Then add the remaining terms: +28x² -1.

Therefore total expression: (-28x² +1) +28x² -1 = 0. So yes! Therefore, if x satisfies 12x³ +14x² -1=0, then it satisfies the sixth-degree equation. Therefore, the cubic equation is a factor of the sixth-degree equation. Therefore, there are other roots, but the original equation is satisfied by a root of the cubic equation. So our original cubic equation is valid. Therefore, the cubic equation is correct.

Therefore, according to this, the minimal |a| + |b| + |c| + |d| is 12 +14 +0 +1=27. But the problem says "the smallest possible value of |a| + |b| + |c| + |d|". Maybe there's a different cubic equation with smaller coefficients?

But the problem says "x satisfies a cubic polynomial of the form...", so the equation 12x³ +14x² -1=0 is a cubic polynomial with integer coefficients. However, maybe there is a different cubic polynomial, perhaps with smaller coefficients, which is equivalent (i.e., scalar multiples) or maybe not? Wait, but if it's a scalar multiple, say k*(12x³ +14x² -1)=0, then the coefficients would be multiplied by k, which would increase |a| + |b| + |c| + |d|. For example, if k=1/2, then coefficients would be 6, 7, 0, -0.5, but since they need to be integers, that's not allowed. Therefore, the minimal coefficients are achieved by the original equation. However, maybe there's a different cubic equation that's not just scalar multiple but different? How can that be? Since the equation was derived through valid steps, and we checked that the cubic is a factor of the sixth-degree equation, but there might be other factors?

But if the original equation is correct, then the cubic equation is the minimal one. However, let me check again if I can find another equation. Wait, but how?

Alternatively, maybe there was an error when solving? Let me check again. From the original equation, we arrived at 12x³ +14x² -1=0. Suppose that there is another approach leading to a different cubic? Let me think.

Alternatively, using more trigonometric identities. Let me suppose that α + β + γ = π. Then maybe considering that α + β = π - γ, and then take sin(α + β) = sin(π - γ) = sin γ. Which is what I did earlier. Then led me to another equation, which after squaring twice gave sixth-degree equation. But since we already have the cubic equation, which is valid, perhaps that is the minimal one. So maybe the answer is 27? But let me check again the problem statement.

Wait, the problem says "the smallest possible value of |a| + |b| + |c| + |d|". Maybe there is another cubic equation with smaller coefficients? Let me think. For example, if the equation can be written as a cubic with smaller coefficients. Let me check if the cubic equation 12x³ +14x² -1=0 can be simplified or if there is a common factor. The coefficients are 12,14,-1. They don't have a common factor except 1. Therefore, this is the minimal polynomial with integer coefficients. Therefore, the answer would be 12 +14 +0 +1=27.

But I need to check again my initial steps. Maybe I made an error when squaring? Because sometimes squaring can introduce extraneous solutions, but the equation we obtained is correct? For example, if x is a solution to the original equation, then it must satisfy the cubic equation. However, the cubic equation may have other solutions which don't satisfy the original equation. However, since we derived the cubic equation through valid steps (assuming that the square roots are positive, which they are in the domain of x). Wait, we have to check the domain of x. Remember that arccos(x), arccos(2x), arccos(3x) must be defined. Therefore:

For arccos(x): x ∈ [-1,1].

For arccos(2x): 2x ∈ [-1,1] ⇒ x ∈ [-1/2, 1/2].

For arccos(3x): 3x ∈ [-1,1] ⇒ x ∈ [-1/3, 1/3].

Therefore, the domain of x is [-1/3, 1/3]. Also, since α + β + γ = π, each of these arccos terms must be between 0 and π, so their sum is π. Therefore, each term is between 0 and π, so x must be positive? Because arccos(x) is between 0 and π, but if x is negative, then arccos(x) is between π/2 and π. However, if x is negative, then 2x and 3x would be more negative, but arccos(2x) and arccos(3x) would require that 2x and 3x are in [-1,1]. However, if x is negative, then 2x and 3x would be negative, so arccos(2x) and arccos(3x) would be in (π/2, π). Therefore, sum of three angles each in (π/2, π) would be larger than 3π/2, which is more than π, but the original equation is sum equals π. Therefore, x must be positive. Therefore, x ∈ (0, 1/3]. Therefore, when solving, we only consider positive roots. So the cubic equation 12x³ +14x² -1=0 has one positive root in (0,1/3], as we saw numerically. Therefore, it is valid.

But just to be thorough, let me check if the cubic equation 12x³ +14x² -1=0 can be factored or transformed into another form with smaller coefficients.

For example, let me write it as:

12x³ +14x² -1 =0.

Is there a way to write this with smaller coefficients? For example, if we can factor it as something like (ax + b)(cx² + dx + e) = 0 with smaller integers?

Let me try to factor it. Suppose it factors as (mx + n)(px² + qx + r) = 0. Then:

m*p =12

m*q + n*p =14

m*r + n*q =0

n*r = -1.

Since n*r = -1, possible integer solutions are n=1, r=-1 or n=-1, r=1.

Let me try n=1, r=-1. Then equations:

m*p=12

m*q + 1*p =14

m*(-1) + 1*q =0 ⇒ -m + q =0 ⇒ q =m.

So substituting q =m into second equation:

m*m + p =14 ⇒ m² + p =14.

But m*p=12. Therefore, m and p are integers such that m*p=12, and m² + p =14.

Let me solve for p =12/m. Then equation becomes m² + 12/m =14. Multiply by m:

m³ +12 =14m ⇒ m³ -14m +12=0.

Looking for integer solutions for m. Possible divisors of 12: ±1, ±2, ±3, ±4, ±6, ±12.

Testing m=1: 1 -14 +12= -1≠0.

m=2: 8 -28 +12= -8≠0.

m=3: 27 -42 +12= -3≠0.

m=4: 64 -56 +12=20≠0.

m= -1: -1 +14 +12=25≠0.

m=-2: -8 +28 +12=32≠0.

No integer solutions. Therefore, this case doesn't work.

Now try n=-1, r=1. Then equations:

m*p =12

m*q + (-1)*p =14 ⇒ m*q -p =14.

m*r + n*q = m*1 + (-1)*q = m - q =0 ⇒ m = q.

Therefore, since m=q, substitute into second equation: m*m -p =14 ⇒ m² -p =14.

But m*p=12 ⇒ p=12/m. Therefore, m² - 12/m =14 ⇒ Multiply by m: m³ -12 =14m ⇒ m³ -14m -12=0.

Looking for integer roots. Divisors of 12: ±1, ±2, ±3, ±4, ±6, ±12.

m=1: 1 -14 -12 = -25 ≠0.

m=2: 8 -28 -12 = -32 ≠0.

m=3: 27 -42 -12 = -27 ≠0.

m=4: 64 -56 -12 = -4 ≠0.

m= -2: -8 +28 -12=8≠0.

m=-3: -27 +42 -12=3≠0.

m=-1: -1 +14 -12=1≠0.

No integer solutions. Therefore, the cubic does not factor with integer coefficients in this way. Therefore, the minimal polynomial is indeed 12x³ +14x² -1. Therefore, the sum |a| + |b| + |c| + |d| is 12 +14 +0 +1=27.

But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|". But maybe there's a different approach where you get a different cubic equation with smaller coefficients? Let me think again.

Wait, another thought. Maybe I made an error in squaring steps, which could lead to another equation, but perhaps combining the equations?

Alternatively, maybe using more identities. For example, let me consider the original equation:

α + β + γ = π. Let me use the identity for three angles summing to π. Maybe express in terms of tangent?

Alternatively, think of α + β = π - γ, and then take tan of both sides? Let me try that.

tan(α + β) = tan(π - γ) = -tan γ.

Compute tan(α + β) = [tan α + tan β]/[1 - tan α tan β] = -tan γ.

But tan α = sqrt(1 -x²)/x? Wait, tan α = sin α / cos α = sqrt(1 -x²)/x. But α = arccos(x), so tan α = sqrt(1 -x²)/x, but only if x > 0, which it is. Similarly tan β = sqrt(1 -4x²)/(2x). tan γ = sqrt(1 -9x²)/(3x). Therefore:

[sqrt(1 -x²)/x + sqrt(1 -4x²)/(2x)] / [1 - (sqrt(1 -x²)/x)(sqrt(1 -4x²)/(2x))] = -sqrt(1 -9x²)/(3x).

Multiply numerator and denominator:

Numerator: [sqrt(1 -x²)/x + sqrt(1 -4x²)/(2x)] = [2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x).

Denominator: 1 - [sqrt{(1 -x²)(1 -4x²)}]/(2x²).

Therefore, equation becomes:

[2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x) divided by [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)] equals -sqrt(1 -9x²)/(3x).

This looks very complicated. Maybe cross-multiplying? Let me denote:

Let me write:

[2 sqrt(1 -x²) + sqrt(1 -4x²)] / (2x) = [ -sqrt(1 -9x²)/(3x) ] * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

Multiply both sides by 2x:

2 sqrt(1 -x²) + sqrt(1 -4x²) = [ -sqrt(1 -9x²)/3x * 2x ] * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

Wait, simplifying the right side:

- sqrt(1 -9x²)/3x * 2x = -2 sqrt(1 -9x²)/3.

Then multiplied by [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

So:

Left side: 2 sqrt(1 -x²) + sqrt(1 -4x²).

Right side: -2 sqrt(1 -9x²)/3 * [1 - sqrt{(1 -x²)(1 -4x²)}/(2x²)].

This seems too complex to solve without squaring again. Perhaps this approach is not helpful. Therefore, returning to the previous result.

Given that we derived the cubic equation 12x³ +14x² -1=0, and checked it through substitution and factoring attempts, and given that it's the minimal polynomial, maybe the answer is 27. However, I need to check if maybe I can scale the equation down. For example, if there's a mistake in coefficients?

Wait, let me check my algebra again when I squared the first time. Original equation after squaring was:

(1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -5x² +4x⁴.

Right side: 4x^4 + 12x³ +9x².

Subtracting 4x^4 from both sides:

1 -5x² = 12x³ +9x² ⇒ 1 -5x² -9x² -12x³ = 0 ⇒ 1 -14x² -12x³ =0 ⇒ -12x³ -14x² +1=0 ⇒ 12x³ +14x² -1=0. Yes, correct. Therefore, the equation is correct. Therefore, the coefficients are correct.

But let me check for typos. For example, is the original equation arccos(x) + arccos(2x) + arccos(3x) = π? Yes. So the steps are correct.

Alternatively, perhaps the problem is expecting a different cubic? Maybe with different coefficients?

Alternatively, maybe there is an error in the sign when expanding? Let me check again:

Original equation after taking cosine of α + β = π - γ:

cos(α + β) = -cos γ.

Which is cos α cos β - sin α sin β = -cos γ.

Substituting cos α =x, cos β =2x, cos γ =3x:

x*2x - sin α sin β = -3x ⇒ 2x² - sin α sin β = -3x ⇒ sin α sin β = 2x² +3x. Correct.

Then sin α sin β = sqrt(1 -x²) * sqrt(1 -4x²) = 2x² +3x. Then squared both sides:

(1 -x²)(1 -4x²) = (2x² +3x)^2. Which is correct. Then expanding:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 + 12x³ +9x².

Subtract 4x^4: 1 -5x² = 12x³ +9x² ⇒ 1 -5x² -9x² -12x³ = 1 -14x² -12x³ =0 ⇒ -12x³ -14x² +1=0 ⇒ 12x³ +14x² -1=0. Yes, correct.

Therefore, I think this is correct. So the answer should be 27. However, let me check if there are other possible solutions or if I missed something.

Wait, but in the problem statement, it says "x satisfies a cubic polynomial...". So perhaps there is another cubic polynomial which is satisfied by x, but with smaller coefficients? For example, if I consider that my equation 12x³ +14x² -1=0 might be equivalent to another cubic with smaller coefficients via some transformation? For example, maybe substituting y = kx or something?

Alternatively, suppose someone claims that the minimal sum is 27, but maybe there is another equation?

Alternatively, let me check if the cubic equation can be written as 12x³ +14x² -1=0 or maybe multiplied by something else?

But since we proved that this is the minimal polynomial, I think that it's safe to go with 27. However, to confirm, I can check online if similar problems have been solved. Wait, but since this is a thought process simulation, I can't actually check. However, I can recall that sometimes problems like this result in equations with smaller coefficients if approached differently, but given the steps I did are correct and verified, I think 27 is the answer.

But just to make sure, let me check with the approximate root. Earlier, I estimated that the root is between 0.24 and 0.25. Let me compute the value of 12x³ +14x² -1 at x≈0.245. As I tried earlier, it was approximately 0.016. Let me try with a better approximation. Let me use Newton-Raphson method to approximate the root.

Let me denote f(x) = 12x³ +14x² -1. We want to find x where f(x)=0.

We have f(0.24)= -0.0277 and f(0.25)=0.0625. Let me take x₀=0.245. f(0.245)= approx. Let me compute more accurately.

x=0.245

x² = 0.245² = 0.060025

x³ = 0.245 * 0.060025 ≈ 0.014706

Then:

12x³ ≈12 * 0.014706 ≈ 0.17647

14x² ≈14 * 0.060025 ≈ 0.84035

Sum: 0.17647 + 0.84035 ≈ 1.01682; subtract 1: ≈0.01682. So f(x)=0.01682 at x=0.245.

f'(x)=36x² +28x. At x=0.245:

36*(0.060025) +28*(0.245) ≈ 2.1609 + 6.86 ≈ 9.0209.

Next iteration: x₁ = x₀ - f(x₀)/f'(x₀) ≈ 0.245 - 0.01682 / 9.0209 ≈ 0.245 - 0.001865 ≈ 0.243135.

Compute f(0.243135):

x ≈0.2431

x² ≈ (0.2431)^2 ≈ 0.0591

x³ ≈ 0.2431 * 0.0591 ≈ 0.01436

Then 12x³ ≈ 0.1723, 14x²≈14*0.0591≈0.8274, total ≈0.1723 +0.8274 -1≈(0.9997) -1≈-0.0003. So f(x)≈-0.0003.

So close to zero. Then f'(x) at x≈0.2431 is 36x² +28x ≈ 36*(0.0591) +28*(0.2431) ≈ 2.1276 +6.8068≈8.9344.

Next iteration: x₁ ≈0.2431 - (-0.0003)/8.9344 ≈0.2431 +0.00003≈0.2431.

Therefore, the root is approximately 0.2431. Let me check the original equation with this value. Let me compute arccos(x) + arccos(2x) + arccos(3x) where x≈0.2431.

Compute x≈0.2431, 2x≈0.4862, 3x≈0.7293.

Compute arccos(0.2431): let me approximate. Let me recall that arccos(0.25)≈1.318 radians. Since 0.2431 is slightly less than 0.25, arccos(0.2431) will be slightly more than 1.318. Let me use calculator-like estimation.

Let me recall that cos(1.33) ≈ cos(1.33). Since π/2 ≈1.5708, so 1.33 is less than π/2, so it's in the first quadrant.

Alternatively, use a Taylor series? Maybe not. Alternatively, use linear approximation around x=0.25. Let me denote f(x)=arccos(x). At x=0.25, f(x)=1.3181. The derivative f’(x)= -1/sqrt(1 -x²). At x=0.25, f’(x)= -1/sqrt(1 -0.0625)= -1/sqrt(0.9375)≈-1/0.9682≈-1.033. Therefore, for x=0.2431, which is Δx = -0.0069 from 0.25, approximate f(x) ≈ f(0.25) + f’(0.25)*Δx ≈ 1.3181 + (-1.033)*(-0.0069) ≈ 1.3181 + 0.0071 ≈1.3252 radians.

Similarly, arccos(2x)=arccos(0.4862). Let me approximate. arccos(0.5)=π/3≈1.0472. 0.4862 is less than 0.5, so arccos(0.4862) is slightly more than π/3. Let me compute f(x)=arccos(x) at x=0.4862. Let me approximate using derivative. Let me take x=0.5, f(x)=π/3≈1.0472. f’(x)= -1/sqrt(1 -0.25)= -1/sqrt(0.75)≈-1.1547. Δx= -0.0138 (since 0.5 -0.4862=0.0138). So f(x)≈1.0472 + (-1.1547)*(-0.0138)≈1.0472 + 0.0159≈1.0631 radians.

Similarly, arccos(3x)=arccos(0.7293). Let me approximate. arccos(0.7)= approx 0.7954 radians. arccos(0.75)=approx 0.7227 radians. Since 0.7293 is between 0.7 and 0.75. Let me approximate. Let me take x=0.7293. Let me linear approximate between x=0.7 and x=0.75. Let me recall that cos(0.7954)=0.7, cos(0.7227)=0.75. So we need to find arccos(0.7293). Let me denote y=arccos(0.7293). Let me take two points:

At x1=0.7, y1=0.7954.

At x2=0.75, y2=0.7227.

We can approximate y using linear interpolation. The value 0.7293 is between 0.7 and 0.75. Let me compute the fraction:

Δx=0.75 -0.7=0.05. The value 0.7293 -0.7=0.0293. Therefore, fraction =0.0293 /0.05≈0.586. So approximate y≈y1 -0.586*(y1 - y2)= 0.7954 -0.586*(0.7954 -0.7227)= 0.7954 -0.586*(0.0727)≈0.7954 -0.0426≈0.7528 radians.

Alternatively, better to use derivative. Let me take x=0.75, y=arccos(0.75)=0.7227. The derivative f’(x)= -1/sqrt(1 -x²). At x=0.75, f’(x)= -1/sqrt(1 -0.5625)= -1/sqrt(0.4375)= -1/0.6614≈-1.511. So if we take x=0.7293, which is Δx=0.7293 -0.75= -0.0207 from x=0.75. Therefore, approximate y≈f(0.75) + f’(0.75)*Δx≈0.7227 -1.511*(-0.0207)≈0.7227 +0.0312≈0.7539 radians.

So sum of the three arccos terms≈1.3252 +1.0631 +0.7539≈3.1422 radians, which is approximately π≈3.1416. Close enough, considering the approximations. Therefore, the root x≈0.2431 satisfies the original equation, and satisfies the cubic equation 12x³ +14x² -1=0. Therefore, the equation is correct.

Hence, the minimal sum is 12 +14 +0 +1=27. Therefore, the answer should be 27.

**Final Answer**
\boxed{27}

To solve the equation:

$$
\arccos x + \arccos 2x + \arccos 3x = \pi,
$$

we define:

$$
\alpha = \arccos x,\quad \beta = \arccos 2x,\quad \gamma = \arccos 3x.
$$

Hence, we have:

$$
\alpha + \beta + \gamma = \pi.
$$

---

### Step 1: Use a Trigonometric Identity

We use the identity:

$$
\alpha + \beta = \pi - \gamma.
$$

Taking the cosine of both sides gives:

$$
\cos(\alpha + \beta) = \cos(\pi - \gamma) = -\cos \gamma.
$$

Now apply the cosine addition formula:

$$
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.
$$

Using the definitions:

$$
\cos \alpha = x,\quad \cos \beta = 2x,\quad \cos \gamma = 3x.
$$

So we get:

$$
\cos(\alpha + \beta) = x \cdot 2x - \sin \alpha \cdot \sin \beta = 2x^2 - \sin \alpha \cdot \sin \beta.
$$

Thus:

$$
2x^2 - \sin \alpha \cdot \sin \beta = -3x.
$$

Which gives:

$$
\sin \alpha \cdot \sin \beta = 2x^2 + 3x.
$$

---

### Step 2: Use Pythagorean Identities

Recall:

$$
\sin \alpha = \sqrt{1 - x^2},\quad \sin \beta = \sqrt{1 - 4x^2}.
$$

Substituting into the equation:

$$
\sqrt{1 - x^2} \cdot \sqrt{1 - 4x^2} = 2x^2 + 3x.
$$

Now square both sides:

$$
(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2.
$$

Expand both sides:

- Left-hand side:

$$
(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4.
$$

- Right-hand side:

$$
(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2.
$$

Equating both:

$$
1 - 5x^2 + 4x^4 = 4x^4 + 12x^3 + 9x^2.
$$

Subtract $ 4x^4 $ from both sides:

$$
1 - 5x^2 = 12x^3 + 9x^2.
$$

Rearrange:

$$
0 = 12x^3 + 14x^2 - 1.
$$

Hence, the equation is:

$$
12x^3 + 14x^2 - 1 = 0.
$$

---

### Step 3: Analyze the Result

We are told that $ x $ satisfies a cubic polynomial of the form:

$$
ax^3 + bx^2 + cx + d = 0,
$$

with integer coefficients and $ a \ne 0 $. Here, we obtain:

$$
a = 12,\quad b = 14,\quad c = 0,\quad d = -1.
$$

Therefore, the sum of the absolute values is:

$$
|a| + |b| + |c| + |d| = 12 + 14 + 0 + 1 = 27.
$$

---

### Step 4: Confirm Minimal Polynomial

We explored whether the cubic could be simplified (e.g., factored or scaled), but it is irreducible over the integers. Attempting rational root tests and factoring into integer polynomials did not reveal smaller coefficients, which confirms that $ 12x^3 + 14x^2 - 1 $ is indeed the minimal cubic polynomial with integer coefficients.

---

### Final Answer

$$
\boxed{27}
$$

Solution:
From the equation $\arccos x + \arccos 2x + \arccos 3x = \pi,$ $\arccos x + \arccos 2x = \pi - \arccos 3x,$ so
\[\cos (\arccos x + \arccos 2x) = \cos (\pi - \arccos 3x).\]From the angle addition formula, the left-hand side becomes
\begin{align*}
\cos (\arccos x + \arccos 2x) &= \cos (\arccos x) \cos (\arccos 2x) - \sin (\arccos x) \sin (\arccos 2x) \\
&= (x)(2x) - (\sqrt{1 - x^2})(\sqrt{1 - 4x^2}) \\
&= 2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)}.
\end{align*}The right-hand side becomes
\[\cos (\pi - \arccos 3x) = -\cos (\arccos 3x) = -3x,\]so
\[2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)} = -3x.\]Then $\sqrt{(1 - x^2)(1 - 4x^2)} = 2x^2 + 3x.$  Squaring both sides, we get
\[(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2.\]This simplifies to $12x^3 + 14x^2 - 1 = 0.$  Thus, the smallest possible value of $|a| + |b| + |c| + |d|$ is $12 + 14 + 0 + 1 = \boxed{27}.$