Okay, so I need to solve the equation arccos x + arccos 2x + arccos 3x = π, and then find the cubic polynomial that x satisfies. Finally, I have to find the smallest possible value of the sum of the absolute values of the coefficients. Hmm, let's start by understanding the problem step by step.

First, arccos is the inverse cosine function, which has a domain of [-1, 1]. So, the arguments of arccos must be within this interval. That means for each term:

1. arccos x: x must be between -1 and 1.
2. arccos 2x: 2x must be between -1 and 1, so x must be between -0.5 and 0.5.
3. arccos 3x: 3x must be between -1 and 1, so x must be between -1/3 and 1/3.

Therefore, the intersection of these domains is x ∈ [-1/3, 1/3]. So, x must be in this interval for the equation to be valid. That's a crucial point because it restricts the possible values of x.

Now, the equation is arccos x + arccos 2x + arccos 3x = π. Let's denote:

Let’s set θ = arccos x, φ = arccos 2x, and ψ = arccos 3x. Then, the equation becomes θ + φ + ψ = π. Since all arccos terms are angles in [0, π], their sum being π suggests that one of them is π minus the sum of the other two. Maybe we can use trigonometric identities or properties of arccos to relate these angles.

I recall that arccos a + arccos b = arccos [ab - sqrt(1 - a²)sqrt(1 - b²)] when certain conditions are met. But here, we have three terms. Maybe I can combine two of them first and then deal with the third.

Alternatively, since θ + φ + ψ = π, perhaps we can take cosine of both sides. Let me try that.

cos(θ + φ + ψ) = cos π = -1.

Using the cosine addition formula: cos(θ + φ + ψ) = cos θ cos φ cos ψ - cos θ sin φ sin ψ - sin θ cos φ sin ψ - sin θ sin φ cos ψ.

But this seems complicated. Maybe there's a smarter way. Let me think.

Alternatively, since θ + φ + ψ = π, then ψ = π - θ - φ. Then, we can take cosine of both sides:

cos ψ = cos(π - θ - φ) = -cos(θ + φ).

But cos ψ is also equal to 3x because ψ = arccos 3x. Similarly, cos(θ + φ) can be expanded as cos θ cos φ - sin θ sin φ. So:

3x = -[cos θ cos φ - sin θ sin φ] = -cos θ cos φ + sin θ sin φ.

So, 3x = -cos θ cos φ + sin θ sin φ.

But cos θ is x, cos φ is 2x, so substituting:

3x = -x*(2x) + sin θ sin φ.

Which simplifies to 3x = -2x² + sin θ sin φ.

So, sin θ sin φ = 3x + 2x².

Now, sin θ and sin φ can be expressed in terms of x. Since θ = arccos x, sin θ = sqrt(1 - x²). Similarly, sin φ = sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore:

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

Let me square both sides to eliminate the square roots:

(1 - x²)(1 - 4x²) = (3x + 2x²)^2.

Expanding both sides:

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

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

So, set left side equal to right side:

1 - 5x² + 4x⁴ = 9x² + 12x³ + 4x⁴.

Subtract left side from both sides:

0 = 9x² + 12x³ + 4x⁴ - (1 - 5x² + 4x⁴) = 9x² + 12x³ + 4x⁴ -1 +5x² -4x⁴.

Simplify:

0 = (9x² +5x²) +12x³ + (4x⁴ -4x⁴) -1 =14x² +12x³ -1.

Thus, 12x³ +14x² -1 =0.

Hmm, so we get 12x³ +14x² -1=0. Let me check if this is correct.

Wait, when I squared both sides, I might have introduced extraneous solutions, so I need to verify the solutions later.

But before that, let me check the coefficients:

Original equation after squaring: (1 - x²)(1 -4x²) = (3x +2x²)^2.

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

Right side: 9x² +12x³ +4x⁴.

So, 1 -5x² +4x⁴ =9x² +12x³ +4x⁴.

Subtract left side:

0=9x² +12x³ +4x⁴ -1 +5x² -4x⁴=14x² +12x³ -1. So yes, 12x³ +14x² -1=0.

But the problem states that the cubic polynomial should be ax³ +bx² +cx +d=0 with integer coefficients, a≠0. So, 12x³ +14x² -1=0 is a cubic with a=12, b=14, c=0, d=-1. Therefore, the sum |a| + |b| + |c| + |d| is 12+14+0+1=27.

But maybe there's a simpler cubic? Let me check if this cubic actually has a solution in the domain x ∈ [-1/3, 1/3]. Let's test 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≠0.

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≠0.

So, x=±1/3 are not roots. Let's try x=0: 0 +0 -1=-1≠0. x=1:12+14-1=25≠0. So, maybe this cubic is correct, but perhaps there's another equation. Wait, let's check my steps again.

Starting from arccos x + arccos 2x + arccos 3x = π.

I set 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(θ + φ).

But cos ψ = 3x, cos(θ + φ) = cos θ cos φ - sin θ sin φ.

Therefore:

3x = -[cos θ cos φ - sin θ sin φ] = -cos θ cos φ + sin θ sin φ.

Substituting cos θ = x, cos φ = 2x, sin θ = sqrt(1 -x²), sin φ = sqrt(1 -4x²):

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

So, 3x = -2x² + sqrt{(1 -x²)(1 -4x²)}.

Then, sqrt{(1 -x²)(1 -4x²)} = 3x + 2x².

Squaring both sides:

(1 -x²)(1 -4x²) = (3x +2x²)^2.

Expanding both sides gives 1 -5x² +4x⁴ =9x² +12x³ +4x⁴.

Subtracting left side from right side:

0=12x³ +14x² -1.

So, yes, that's correct. Therefore, the cubic is 12x³ +14x² -1=0.

But maybe there is a lower-degree polynomial that x satisfies? Let me check if this cubic can be factored or simplified.

Looking at 12x³ +14x² -1. Let's try rational roots. Possible rational roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12.

Testing x=1: 12 +14 -1=25≠0.

x=-1: -12 +14 -1=1≠0.

x=1/2: 12*(1/8) +14*(1/4) -1=1.5 +3.5 -1=4≠0.

x=-1/2: 12*(-1/8)+14*(1/4)-1=-1.5 +3.5 -1=1≠0.

x=1/3: 12*(1/27)+14*(1/9)-1≈0.444 +1.555 -1≈1≠0.

x=-1/3: 12*(-1/27)+14*(1/9)-1≈-0.444 +1.555 -1≈0.111≠0.

x=1/4: 12*(1/64)+14*(1/16)-1=0.1875 +0.875 -1=0.0625≠0.

x=-1/4: 12*(-1/64)+14*(1/16)-1≈-0.1875 +0.875 -1≈-0.3125≠0.

x=1/6: 12*(1/216)+14*(1/36)-1≈0.055 +0.388 -1≈-0.557≠0.

x=-1/6: 12*(-1/216)+14*(1/36)-1≈-0.055 +0.388 -1≈-0.667≠0.

x=1/12: 12*(1/1728)+14*(1/144)-1≈0.0069 +0.0972 -1≈-0.8959≠0.

x=-1/12: 12*(-1/1728)+14*(1/144)-1≈-0.0069 +0.0972 -1≈-0.9099≠0.

So, no rational roots. Therefore, the cubic is irreducible over rationals, so the minimal polynomial is indeed 12x³ +14x² -1=0. Therefore, the coefficients are a=12, b=14, c=0, d=-1.

Hence, the sum |12| + |14| + |0| + |-1| =12+14+0+1=27.

But wait, the problem says "satisfies a cubic polynomial of the form ax³ +bx² +cx +d=0", so we need to confirm if this is indeed the minimal polynomial, or if there's a lower-degree polynomial with integer coefficients. However, since x is in the interval [-1/3, 1/3], we need to check if x can satisfy a quadratic equation.

Suppose there exists a quadratic equation with integer coefficients that x satisfies. Let's see.

Suppose ax³ +bx² +cx +d=0 is reducible to a quadratic. Then, it must factor as (x - r)(quadratic) where r is a root. But since the cubic is irreducible, as we saw, there are no rational roots, so it can't be factored into a linear term and a quadratic with integer coefficients. Therefore, the minimal polynomial is indeed cubic, so the answer is 27.

But let me verify once again to make sure I didn't make any miscalculations.

Starting again, the equation:

arccos x + arccos 2x + arccos 3x = π.

Let me pick a value of x in the domain [-1/3, 1/3] and check if the equation holds. For example, x=0. Let's see:

arccos 0 = π/2,

arccos 0 = π/2,

arccos 0 = π/2. So, sum is 3*(π/2) = 3π/2 ≠ π. So, x=0 is not a solution.

x=1/3:

arccos(1/3) + arccos(2/3) + arccos(1) = arccos(1/3) + arccos(2/3) + 0. Let's compute approximate values:

arccos(1/3) ≈ 1.23096 radians,

arccos(2/3) ≈ 0.84106 radians,

Sum ≈1.23096 +0.84106≈2.072 radians, which is less than π≈3.1416. So, sum is less than π. So, x=1/3 is too big.

x=1/4:

arccos(1/4)≈1.318 radians,

arccos(1/2)=π/3≈1.047 radians,

arccos(3/4)≈0.7227 radians,

Sum≈1.318 +1.047 +0.7227≈3.087 radians, which is still less than π (≈3.1416). So, close. Maybe x=1/4 is a solution?

But according to our earlier cubic equation, x=1/4 gives 12*(1/4)^3 +14*(1/4)^2 -1=12*(1/64)+14*(1/16)-1≈0.1875 +0.875 -1≈0.0625≠0. So, x=1/4 is not a root. So, maybe the actual solution is near x=1/4, but not exactly 1/4.

Alternatively, maybe the cubic is correct.

Alternatively, perhaps I made a mistake in the trigonometric identity step. Let me check again.

Starting from θ + φ + ψ = π.

Then, ψ = π - θ - φ.

So, cos ψ = cos(π - θ - φ) = -cos(θ + φ).

But cos(θ + φ) = cos θ cos φ - sin θ sin φ. So, cos ψ = -[cos θ cos φ - sin θ sin φ] = -cos θ cos φ + sin θ sin φ.

But cos ψ is 3x, so:

3x = -cos θ cos φ + sin θ sin φ.

Substituting cos θ = x, cos φ =2x:

3x = -x*(2x) + sin θ sin φ.

So, 3x = -2x² + sin θ sin φ.

Therefore, sin θ sin φ = 3x +2x².

But sin θ = sqrt(1 -x²), sin φ = sqrt(1 - (2x)^2) = sqrt(1 -4x²).

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

Squaring both sides:

(1 -x²)(1 -4x²) = (3x +2x²)^2.

Which gives 1 -5x² +4x⁴ =9x² +12x³ +4x⁴.

Subtracting left side:

0=12x³ +14x² -1.

Yes, same result. So, the cubic is correct.

Therefore, the minimal polynomial is 12x³ +14x² -1=0, so the sum of absolute values is 12+14+0+1=27.

But the problem says "the cubic polynomial of the form ax³ +bx² +cx +d=0", so the coefficients are 12,14,0,-1. Therefore, the sum is 27. But the problem asks for the smallest possible value of |a| + |b| + |c| + |d|. Since 27 is the sum here, but maybe there's a multiple of this equation with smaller coefficients?

If we multiply the equation by a constant, the coefficients would change, but since we need integer coefficients, the minimal coefficients would be when the polynomial is primitive (gcd of coefficients is 1). Since 12,14,0,-1 have gcd 1, so this is the minimal polynomial. Therefore, the sum is indeed 27.

Therefore, the answer is 27. So, the final answer is \boxed{27}.

**Final Answer**
\boxed{27}
Therefore, the smallest possible value of $|a| + |b| + |c| + |d|$ is $\boxed{27}$.
</think>