Explanation:
Okay, so I need to solve this equation: arccos x + arccos 2x + arccos 3x = π. And then find the cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d|. Alright, let me start by recalling some trigonometric identities and properties of inverse cosine functions.

First, I know that the range of arccos is [0, π], so each term arccos x, arccos 2x, arccos 3x must be in that interval. Therefore, the arguments of arccos must be in [-1, 1]. So, for arccos x to be defined, x must be in [-1, 1]. Similarly, 2x must be in [-1, 1], so x ∈ [-0.5, 0.5]. Similarly, 3x must be in [-1, 1], so x ∈ [-1/3, 1/3]. Therefore, combining all these, x must be in [-1/3, 1/3]. So x is between -1/3 and 1/3. But since arccos is decreasing, and the sum of three arccos terms equals π, which is a positive number. Let me think about the possible values of x. If x is negative, then arccos x would be in (π/2, π), arccos 2x would be in (π/2, π) if 2x is negative, same with arccos 3x. But adding three angles each greater than π/2 would give more than 3π/2, which is more than π. So maybe x is positive? Let me check.

If x is positive, then arccos x, arccos 2x, arccos 3x are all in [0, π/2], since their arguments are positive and less than or equal to 1. Wait, but 3x is up to 1 when x is 1/3. So arccos(3x) would be arccos(1) = 0 when x = 1/3. So if x is positive, then arccos x is between 0 and π/2, arccos 2x is between 0 and π/2, arccos 3x is between 0 and π/2. So adding three angles between 0 and π/2. The maximum sum would be 3*(π/2) = 3π/2, but the equation says the sum is π. So maybe x is positive? Let me check if x can be negative. Suppose x is negative. Then arccos x is in (π/2, π), arccos 2x is in (π/2, π) if 2x is between -1 and 0, same with arccos 3x. So adding three angles each between π/2 and π would give a sum between 3π/2 and 3π. But the equation says the sum is π, which is less than 3π/2. Therefore, x cannot be negative. Therefore, x must be positive. Therefore, x ∈ (0, 1/3].

So x is between 0 and 1/3. So that's helpful. So we can focus on x in (0, 1/3].

Now, the equation is arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. So maybe I can use some trigonometric identities here. Let me think. If I take cos of both sides? Let me try that.

But cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) might be complicated. Alternatively, maybe I can use the identity for cos(α + β) = cos π - γ. Wait, since α + β + γ = π, then α + β = π - γ. Therefore, cos(α + β) = cos(π - γ) = -cos γ. So cos(α + β) = -cos γ. Let me compute cos(α + β) using the formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos(α + β) = -cos γ. Therefore:

cos α cos β - sin α sin β = -cos γ.

But cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x.

So 2x² - sin α sin β = -3x.

Therefore, sin α sin β = 2x² + 3x.

Hmm. Now, I need to find expressions for sin α and sin β. Since α = arccos x, then sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore:

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

So that's an equation in x. Let me square both sides to eliminate the square roots. But before that, let me note that squaring can introduce extraneous solutions, so I need to check later.

So squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2.

Let me compute both sides.

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

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

Therefore, equation becomes:

1 - 5x² + 4x^4 = 4x^4 + 12x^3 + 9x².

Subtracting left side from both sides:

0 = 4x^4 + 12x^3 + 9x² - (1 - 5x² + 4x^4) = 4x^4 + 12x^3 + 9x² -1 + 5x² -4x^4.

Simplify:

4x^4 -4x^4 cancels. Then 12x^3 + 9x² +5x² -1 = 12x^3 +14x² -1 = 0.

Therefore, the equation reduces to 12x³ +14x² -1 = 0. Wait, but hold on. Let me check my algebra again.

Original equation after squaring:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

So bringing left side to the right:

0 = 4x^4 +12x^3 +9x² -1 +5x² -4x^4.

Wait, no. Wait, original equation is left side equals right side. So:

Left side = right side => 1 -5x² +4x^4 = 4x^4 +12x^3 +9x².

Subtract left side from both sides:

0 = 4x^4 +12x^3 +9x² - (1 -5x² +4x^4) = 4x^4 +12x^3 +9x² -1 +5x² -4x^4.

Yes, so 4x^4 -4x^4 cancels. Then 12x^3 + (9x² +5x²) -1 = 12x^3 +14x² -1 = 0. Therefore, 12x³ +14x² -1 = 0. So that's a cubic equation. But wait, the problem says that x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0. So here, we have 12x³ +14x² +0x -1 = 0. Therefore, coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a cubic polynomial with smaller coefficients? Or maybe I made a mistake here?

Wait, but let me check if this equation is correct. Let me verify my steps again.

Starting from α + β + γ = π. Then α + β = π - γ. Then cos(α + β) = cos(π - γ) = -cos γ. Then expanding cos(α + β) = cos α cos β - sin α sin β = -cos γ. Then substituting cos α = x, cos β = 2x, cos γ = 3x. Therefore:

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

(1 - x²)(1 - 4x²) = (2x² + 3x)^2. Then expanding both sides:

Left side: 1 - 5x² + 4x^4.

Right side: 4x^4 + 12x^3 + 9x².

Subtracting left side from right side:

(4x^4 + 12x^3 + 9x²) - (1 -5x² +4x^4) = 12x^3 + 14x² -1 = 0. So that's correct. Therefore, the equation is 12x³ +14x² -1 = 0. Therefore, coefficients are 12, 14, 0, -1. So |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 is another way to get a different cubic equation with smaller coefficients? Or maybe I need to check if this equation is correct?

Wait, but maybe there are extraneous solutions introduced by squaring. So perhaps the original equation has solutions that don't satisfy the squared equation? Let me check.

So suppose x is a solution to 12x³ +14x² -1 = 0. Let me check if it satisfies the original equation. Let me try to find approximate solutions. Let me consider x in (0, 1/3]. Let me compute f(x) = 12x³ +14x² -1. At x=0, f(0) = -1. At x=1/3, f(1/3) = 12*(1/27) +14*(1/9) -1 = 12/27 +14/9 -1 = 4/9 +14/9 -1 = 18/9 -1 = 2 -1 = 1. Therefore, since f(0) = -1 and f(1/3) = 1, by Intermediate Value Theorem, there is a root between 0 and 1/3. Therefore, there is a solution. Let me check if that solution satisfies the original equation. But how?

Alternatively, maybe I can check if the equation 12x³ +14x² -1 = 0 is the minimal polynomial. But maybe there is a mistake in the derivation? Let me think again.

Alternatively, maybe I can use another approach. Let me try to use more identities. Let me consider that α + β + γ = π. So maybe I can take another pair of angles. For example, take α + γ = π - β. Then take cos(α + γ) = cos(π - β) = -cos β. So cos(α + γ) = -cos β. Let me compute cos(α + γ) = cos α cos γ - sin α sin γ. Which would be x*(3x) - sin α sin γ = 3x² - sin α sin γ = -cos β = -2x. Therefore:

3x² - sin α sin γ = -2x => sin α sin γ = 3x² + 2x.

Similarly, earlier we had sin α sin β = 2x² + 3x. So now we have two equations:

sin α sin β = 2x² + 3x,

sin α sin γ = 3x² + 2x.

Let me write them down:

Equation 1: sin α sin β = 2x² + 3x,

Equation 2: sin α sin γ = 3x² + 2x.

Also, we can do similar for β + γ = π - α. Let me try that. cos(β + γ) = cos(π - α) = -cos α = -x. Then cos(β + γ) = cos β cos γ - sin β sin γ = 2x * 3x - sin β sin γ = 6x² - sin β sin γ = -x. Therefore:

6x² - sin β sin γ = -x => sin β sin γ = 6x² + x.

So now we have three equations:

1. sin α sin β = 2x² + 3x,

2. sin α sin γ = 3x² + 2x,

3. sin β sin γ = 6x² + x.

So now, if I denote sin α = s, sin β = t, sin γ = u. Then:

s*t = 2x² + 3x,

s*u = 3x² + 2x,

t*u = 6x² + x.

So we have three equations with three variables s, t, u. Let me try to solve for s, t, u. Let me multiply all three equations together:

(s*t)*(s*u)*(t*u) = (2x² + 3x)(3x² + 2x)(6x² + x).

Left side: s² t² u² = (s t u)^2.

Right side: Let me compute that. Let me first compute (2x² + 3x)(3x² + 2x). Let me compute that first:

(2x² + 3x)(3x² + 2x) = 2x²*3x² + 2x²*2x + 3x*3x² + 3x*2x = 6x^4 + 4x³ + 9x³ + 6x² = 6x^4 + 13x³ + 6x².

Then multiply by (6x² + x):

(6x^4 + 13x³ + 6x²)(6x² + x) = Let me compute term by term:

First, 6x^4 * 6x² = 36x^6,

6x^4 * x = 6x^5,

13x³ * 6x² = 78x^5,

13x³ * x = 13x^4,

6x² * 6x² = 36x^4,

6x² * x = 6x³.

Adding all together:

36x^6 + (6x^5 + 78x^5) + (13x^4 + 36x^4) + 6x³ = 36x^6 + 84x^5 + 49x^4 + 6x³.

Therefore, the right side is 36x^6 + 84x^5 + 49x^4 + 6x³.

On the left side, (s t u)^2. Let me find s t u. Let me note that from equations:

From equation 1: s t = 2x² + 3x,

From equation 2: s u = 3x² + 2x,

From equation 3: t u = 6x² + x.

Let me solve for s, t, u. Let me take the product of equations 1, 2, 3:

(s t)(s u)(t u) = (2x² + 3x)(3x² + 2x)(6x² + x) = (s t u)^2, which we already calculated as 36x^6 + 84x^5 + 49x^4 + 6x³. Therefore, s t u = sqrt(36x^6 + 84x^5 + 49x^4 + 6x³). But maybe we can find s t u another way?

Alternatively, let me solve for s, t, u. Let me take equation 1: s t = 2x² + 3x,

equation 2: s u = 3x² + 2x,

equation 3: t u = 6x² + x.

Let me solve for s, t, u. Let me divide equation 2 by equation 1: (s u)/(s t) = (3x² + 2x)/(2x² + 3x) => u/t = (3x² + 2x)/(2x² + 3x). Therefore, u = t*(3x² + 2x)/(2x² + 3x). Similarly, from equation 3: t u = 6x² + x. Substitute u:

t * [ t*(3x² + 2x)/(2x² + 3x) ] = 6x² + x => t² * (3x² + 2x)/(2x² + 3x) = 6x² + x.

Therefore, t² = (6x² + x) * (2x² + 3x)/(3x² + 2x). Let me compute that:

First, note that 6x² + x = x(6x + 1), and 2x² + 3x = x(2x + 3), and 3x² + 2x = x(3x + 2). Therefore:

t² = [x(6x + 1) * x(2x + 3)] / [x(3x + 2)] = [x²(6x + 1)(2x + 3)] / [x(3x + 2)] = x(6x + 1)(2x + 3)/(3x + 2).

Therefore, t² = x(6x + 1)(2x + 3)/(3x + 2). Similarly, since t = sin β, which is sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore, t² = 1 - 4x². Therefore:

1 - 4x² = x(6x + 1)(2x + 3)/(3x + 2).

Let me compute the right side:

First compute (6x + 1)(2x + 3) = 12x² + 18x + 2x + 3 = 12x² + 20x + 3.

Therefore, right side is x*(12x² + 20x + 3)/(3x + 2). Therefore:

1 - 4x² = [x(12x² + 20x + 3)] / (3x + 2).

Multiply both sides by (3x + 2):

(1 - 4x²)(3x + 2) = x(12x² + 20x + 3).

Let me expand left side:

1*(3x + 2) - 4x²*(3x + 2) = 3x + 2 - 12x³ - 8x².

Right side: 12x³ + 20x² + 3x.

Therefore, equation becomes:

3x + 2 - 12x³ - 8x² = 12x³ + 20x² + 3x.

Subtract left side from both sides:

0 = 12x³ + 20x² + 3x - (3x + 2 - 12x³ - 8x²) = 12x³ + 20x² + 3x -3x -2 +12x³ +8x².

Simplify:

12x³ +12x³ = 24x³,

20x² +8x² = 28x²,

3x -3x = 0,

-2.

Therefore, 24x³ +28x² -2 = 0. Divide both sides by 2: 12x³ +14x² -1 = 0. Which is the same equation as before! So that confirms that the equation is correct. Therefore, the cubic equation is indeed 12x³ +14x² -1 = 0. Therefore, the coefficients are a=12, b=14, c=0, d=-1. Therefore, |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a different cubic equation with smaller coefficients? Or maybe this is the minimal one?

Wait, but maybe the equation can be scaled? For example, if we multiply by some integer, but the problem says "a cubic polynomial of the form ax³ +bx² +cx +d = 0, where a, b, c, and d are integers, and a ≠ 0." So the minimal polynomial is 12x³ +14x² -1, but maybe there is another polynomial with smaller coefficients? For example, if there is a common factor? Let me check if 12x³ +14x² -1 can be factored. Let me try rational roots. By Rational Root Theorem, possible rational roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12. Let me test them.

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*(-1/2)^3 +14*(-1/2)^2 -1 = 12*(-1/8) +14*(1/4) -1 = -1.5 + 3.5 -1 = 1 ≠0.

x=1/3: 12*(1/27) +14*(1/9) -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 = -4/9 +14/9 -1 = 10/9 -1 = 1/9 ≠0.

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 ≠0.

x=-1/4: 12*(-1/4)^3 +14*(-1/4)^2 -1 = 12*(-1/64) +14*(1/16) -1 = -3/16 +7/8 -1 = -3/16 +14/16 -16/16 = (-3 +14 -16)/16 = (-5)/16 ≠0.

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 ≠0.

x=-1/6: 12*(-1/6)^3 +14*(-1/6)^2 -1 = 12*(-1/216) +14*(1/36) -1 = -1/18 +7/18 -1 = 6/18 -1 = 1/3 -1 = -2/3 ≠0.

x=1/12: 12*(1/12)^3 +14*(1/12)^2 -1 = 12*(1/1728) +14*(1/144) -1 = 1/144 + 7/72 -1 = 1/144 + 14/144 -144/144 = (1 +14 -144)/144 = (-129)/144 ≠0.

Similarly for others. Therefore, there are no rational roots. Therefore, the polynomial 12x³ +14x² -1 is irreducible over rationals? Therefore, it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial. Therefore, the coefficients are 12, 14, 0, -1, so the sum is 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." Maybe there is another approach where we get a different cubic equation with smaller coefficients? Let me think.

Alternatively, maybe I made a mistake in assuming that squaring didn't introduce extraneous solutions? Let me check if the equation 12x³ +14x² -1 = 0 has solutions in (0, 1/3]. Let me approximate the root. Let me take f(x) = 12x³ +14x² -1. At x=0.2: f(0.2)=12*(0.008) +14*(0.04) -1 = 0.096 + 0.56 -1 = -0.344. At x=0.25: f(0.25)=12*(0.015625) +14*(0.0625) -1 = 0.1875 + 0.875 -1 = 1.0625 -1 = 0.0625. So between 0.2 and 0.25, f(x) crosses zero. Let me try x=0.23: f(0.23)=12*(0.012167) +14*(0.0529) -1 ≈ 0.146 + 0.7406 -1 ≈ 0.8866 -1 ≈ -0.1134. x=0.24: 12*(0.013824) +14*(0.0576) -1 ≈ 0.1659 + 0.8064 -1 ≈ 0.9723 -1 ≈ -0.0277. x=0.245: 12*(0.245)^3 +14*(0.245)^2 -1. Let me compute 0.245^2 = 0.060025, 0.245^3 ≈ 0.014706. Then 12*0.014706 ≈ 0.1765, 14*0.060025 ≈ 0.84035. Total ≈ 0.1765 + 0.84035 -1 ≈ 1.01685 -1 ≈ 0.01685. So between 0.24 and 0.245, f(x) crosses zero. So the root is approximately 0.242. Let me check if this value satisfies the original equation arccos x + arccos 2x + arccos 3x = π. Let me compute approximate values.

Let me take x ≈ 0.242. Then:

arccos(0.242) ≈ let me compute. Since cos(π/3) = 0.5, cos(π/2) = 0. So arccos(0.242) is between π/3 ≈ 1.047 and π/2 ≈ 1.5708. Let me use calculator-like approximations. Let me recall that cos(1.35 radians) ≈ cos(77.3 degrees) ≈ 0.2225. Wait, 0.242 is larger than 0.2225, so arccos(0.242) is less than 1.35 radians. Let me try 1.3 radians: cos(1.3) ≈ 0.2675. 1.3 radians is about 74.5 degrees. So cos(1.3) ≈ 0.2675. Our x is 0.242, which is less than 0.2675, so arccos(0.242) is greater than 1.3 radians. Let me try 1.33 radians: cos(1.33) ≈ cos(76.2 degrees). Let me compute cos(1.33):

Using Taylor series or calculator? Well, since I don't have a calculator here, but let me approximate. Let me use linear approximation between 1.3 and 1.35. At 1.3, cos(1.3) ≈ 0.2675. At 1.35, cos(1.35) ≈ 0.2225. The difference between 1.3 and 1.35 is 0.05 radians, and the difference in cos is 0.2225 - 0.2675 = -0.045. So slope is -0.045 / 0.05 = -0.9 per radian. So to find x where cos(x) = 0.242, starting from 1.3 where cos(1.3)=0.2675. Let me set up linear approximation:

Let me denote f(x) = cos(x). We have f(1.3) = 0.2675, f'(x) = -sin(x). At x=1.3, sin(1.3) ≈ sin(74.5 degrees) ≈ 0.9613. Therefore, f'(1.3) ≈ -0.9613. So linear approximation:

f(x) ≈ f(1.3) + f'(1.3)(x - 1.3). We want f(x) = 0.242. Therefore:

0.242 ≈ 0.2675 - 0.9613*(x - 1.3)

=> 0.242 - 0.2675 ≈ -0.9613*(x - 1.3)

=> -0.0255 ≈ -0.9613*(x - 1.3)

=> x - 1.3 ≈ 0.0255 / 0.9613 ≈ 0.0265

=> x ≈ 1.3 + 0.0265 ≈ 1.3265 radians.

So arccos(0.242) ≈ 1.3265 radians.

Similarly, arccos(2x) = arccos(0.484). Let me compute that. cos(π/3) = 0.5, so arccos(0.484) is slightly larger than π/3 ≈ 1.047 radians. Let me approximate. Let me take x=1.047 radians, cos(1.047)=0.5. Let me try 1.05 radians: cos(1.05) ≈ cos(60.2 degrees). Let me compute cos(1.05):

Again, using linear approximation. Let me take x=1.047 (π/3), cos(π/3)=0.5. Let me take derivative at π/3: -sin(π/3) ≈ -√3/2 ≈ -0.866. So f(x) = cos(x). Let me approximate cos(1.05):

Let me write 1.05 = π/3 + Δx, where Δx = 1.05 - π/3 ≈ 1.05 - 1.047 ≈ 0.003 radians. Then cos(π/3 + Δx) ≈ cos(π/3) - sin(π/3)*Δx ≈ 0.5 - 0.866*0.003 ≈ 0.5 - 0.0026 ≈ 0.4974. But we need cos(x) = 0.484, which is less than 0.4974. Therefore, need to go further. Let me try Δx = 0.01 radians. Then cos(π/3 + 0.01) ≈ 0.5 - 0.866*0.01 ≈ 0.5 - 0.00866 ≈ 0.4913. Still higher than 0.484. Let me try Δx = 0.015 radians: cos(π/3 + 0.015) ≈ 0.5 - 0.866*0.015 ≈ 0.5 - 0.01299 ≈ 0.487. Closer. Need 0.484. Let me try Δx = 0.017 radians: 0.5 - 0.866*0.017 ≈ 0.5 - 0.0147 ≈ 0.4853. Still higher. Δx=0.018: 0.5 - 0.866*0.018 ≈ 0.5 - 0.0156 ≈ 0.4844. Close. So arccos(0.484) ≈ π/3 + 0.018 ≈ 1.047 + 0.018 ≈ 1.065 radians.

Similarly, arccos(3x) = arccos(0.726). Let me compute that. cos(π/4) ≈ 0.707, so arccos(0.726) is slightly less than π/4 ≈ 0.785 radians. Let me approximate. Let me take x=π/4 ≈ 0.785, cos(π/4)=√2/2 ≈ 0.707. Let me compute cos(0.75 radians): cos(0.75) ≈ 0.7317. Which is higher than 0.726. Let me try 0.76 radians: cos(0.76) ≈ cos(43.5 degrees). Let me use linear approximation. Let me take x=0.75, cos(0.75)=0.7317. Let me take derivative at x=0.75: -sin(0.75) ≈ -sin(43 degrees) ≈ -0.682. So f(x) = cos(x). Let me approximate cos(0.75 + Δx) ≈ 0.7317 - 0.682*Δx. We want cos(x) = 0.726. So:

0.726 ≈ 0.7317 - 0.682*Δx => Δx ≈ (0.7317 - 0.726)/0.682 ≈ 0.0057 / 0.682 ≈ 0.00836 radians. Therefore, arccos(0.726) ≈ 0.75 + 0.00836 ≈ 0.7584 radians.

Therefore, sum of arccos x + arccos 2x + arccos 3x ≈ 1.3265 + 1.065 + 0.7584 ≈ 1.3265 + 1.065 = 2.3915 + 0.7584 ≈ 3.1499 radians. Which is approximately π ≈ 3.1416. So the approximate value is about 3.15, which is slightly larger than π. Therefore, there might be some error due to my approximations. But it's close. Therefore, the root of the equation 12x³ +14x² -1 = 0 is indeed a solution to the original equation. Therefore, the cubic equation is correct.

But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is another cubic equation with smaller coefficients? For example, if we can factor out something? But since the polynomial is irreducible over rationals, as we saw, there are no rational roots, so it cannot be factored into a product of lower-degree polynomials with integer coefficients. Therefore, the minimal polynomial is indeed 12x³ +14x² -1. Therefore, the sum is 12 +14 +0 +1 = 27. However, maybe there is another approach where we get a different cubic equation? Let me think again.

Alternatively, maybe I can use another identity. Let me consider taking the original equation arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me take the first two terms: α + β = π - γ. Then take cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ. Which is what I did before. So that gives me the equation leading to 12x³ +14x² -1 = 0. Alternatively, maybe I can use sine? Let me think. Since α + β + γ = π, then maybe take sine of both sides? Let me try:

sin(α + β + γ) = sin π = 0. But expanding sin(α + β + γ) is complicated. Alternatively, since α + β = π - γ, then sin(α + β) = sin(π - γ) = sin γ. So sin(α + β) = sin γ. Expanding sin(α + β) = sin α cos β + cos α sin β = sin γ. Therefore:

sin α cos β + cos α sin β = sin γ.

We know cos α = x, cos β = 2x, cos γ = 3x. Also, sin α = sqrt(1 - x²), sin β = sqrt(1 - 4x²), sin γ = sqrt(1 - 9x²). Therefore:

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

This seems another equation. Let me write it:

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

This seems more complicated, but maybe squaring both sides? Let me try. Let me denote LHS = 2x sqrt(1 - x²) + x sqrt(1 - 4x²), RHS = sqrt(1 - 9x²). Squaring both sides:

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

Expanding the left side:

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

Compute each term:

First term: 4x²(1 - x²).

Second term: 4x² sqrt{(1 - x²)(1 - 4x²)}.

Third term: x²(1 - 4x²).

Therefore, total left side:

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

Simplify the first two terms:

4x² - 4x^4 + x² - 4x^4 = 5x² - 8x^4.

Therefore, equation becomes:

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

Bring the non-radical terms to the right:

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

Therefore:

sqrt{(1 - x²)(1 - 4x²)} = [1 -14x² +8x^4]/(4x²).

Square both sides again:

(1 - x²)(1 - 4x²) = [ (1 -14x² +8x^4)^2 ] / (16x^4).

Multiply both sides by 16x^4:

16x^4(1 - x²)(1 - 4x²) = (1 -14x² +8x^4)^2.

This seems very complicated, but let me compute both sides.

Left side: 16x^4*(1 - x²)(1 - 4x²). Earlier, we had (1 - x²)(1 - 4x²) = 1 -5x² +4x^4. Therefore, left side is 16x^4*(1 -5x² +4x^4) = 16x^4 -80x^6 +64x^8.

Right side: (1 -14x² +8x^4)^2. Let me expand this:

Let me denote A = 8x^4 -14x² +1. Then A² = (8x^4)^2 + (-14x²)^2 +1^2 + 2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1).

Wait, more systematically:

(8x^4 -14x² +1)^2 = (8x^4)^2 + (-14x²)^2 + (1)^2 + 2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1).

Compute term by term:

(8x^4)^2 = 64x^8,

(-14x²)^2 = 196x^4,

(1)^2 = 1,

Cross terms:

2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1) = 2*(-112x^6 +8x^4 -14x²) = -224x^6 +16x^4 -28x².

Therefore, total right side:

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

Therefore, equation is:

Left side: 16x^4 -80x^6 +64x^8 = Right side: 64x^8 -224x^6 +212x^4 -28x² +1.

Subtract left side from both sides:

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

Simplify:

64x^8 -64x^8 = 0,

-224x^6 +80x^6 = -144x^6,

212x^4 -16x^4 = 196x^4,

-28x²,

+1.

Therefore, equation becomes:

-144x^6 +196x^4 -28x² +1 = 0.

Multiply both sides by -1:

144x^6 -196x^4 +28x² -1 = 0.

This is a sixth-degree equation. However, we already have a cubic equation from before. Let me check if this sixth-degree equation is related to the cubic equation. Let me note that if x satisfies 12x³ +14x² -1 = 0, then maybe substituting x³ = (-14x² +1)/12 into this equation. Let me try.

Let me denote the sixth-degree equation: 144x^6 -196x^4 +28x² -1 = 0. Let me express x^6 in terms of lower degrees. From the cubic equation 12x³ +14x² -1 = 0, we can solve for x³:

x³ = (1 -14x²)/12.

Therefore, x^6 = (x³)^2 = [(1 -14x²)/12]^2 = (1 -28x² +196x^4)/144.

Substitute into the sixth-degree equation:

144x^6 -196x^4 +28x² -1 = 144*(1 -28x² +196x^4)/144 -196x^4 +28x² -1 = (1 -28x² +196x^4) -196x^4 +28x² -1 = 1 -28x² +196x^4 -196x^4 +28x² -1 = 0. Therefore, the sixth-degree equation is satisfied by the roots of the cubic equation. Therefore, the sixth-degree equation is a multiple of the cubic equation? Or maybe the cubic equation is a factor? Let me check.

Let me perform polynomial division: divide 144x^6 -196x^4 +28x² -1 by 12x³ +14x² -1. Let me see.

Let me write the sixth-degree polynomial as 144x^6 + 0x^5 -196x^4 +0x^3 +28x² +0x -1.

Divide by 12x³ +14x² -1.

Let me use polynomial long division.

Divide 144x^6 by 12x³: 12x³. Multiply divisor by 12x³: 12x³*(12x³ +14x² -1) = 144x^6 +168x^5 -12x³.

Subtract this from the dividend:

(144x^6 +0x^5 -196x^4 +0x^3 +28x² +0x -1) - (144x^6 +168x^5 -12x³) = 0x^6 -168x^5 -196x^4 +12x³ +28x² +0x -1.

Next term: -168x^5 divided by 12x³ is -14x². Multiply divisor by -14x²: -14x²*(12x³ +14x² -1) = -168x^5 -196x^4 +14x².

Subtract this from the current dividend:

(-168x^5 -196x^4 +12x³ +28x² +0x -1) - (-168x^5 -196x^4 +14x²) = 0x^5 +0x^4 +12x³ + (28x² -14x²) +0x -1 = 12x³ +14x² -1.

Now, divide 12x³ +14x² -1 by the divisor 12x³ +14x² -1, which is 1. Therefore, the division yields quotient 12x³ -14x² +1, and remainder zero. Therefore, the sixth-degree polynomial factors as (12x³ +14x² -1)(12x³ -14x² +1). Therefore, the sixth-degree equation factors into two cubics. Therefore, the original equation 144x^6 -196x^4 +28x² -1 = 0 is equivalent to (12x³ +14x² -1)(12x³ -14x² +1) = 0. Therefore, the solutions are the roots of 12x³ +14x² -1 = 0 and 12x³ -14x² +1 = 0. However, we need to check which of these cubics have roots in the interval (0, 1/3].

Let me check the second cubic: 12x³ -14x² +1 = 0. Let me check for roots in (0, 1/3]. Let me define f(x) = 12x³ -14x² +1. At x=0: f(0)=1. At x=1/3: f(1/3)=12*(1/27) -14*(1/9) +1 = 4/9 -14/9 +1 = (-10/9) +1 = -1/9. Therefore, since f(0)=1 and f(1/3)=-1/9, there is a root between 0 and 1/3. Therefore, both cubics have roots in (0, 1/3]. However, we need to check which one is the correct one. Let me check with the approximate value we had earlier, x≈0.242. Let me plug into 12x³ +14x² -1 ≈ 0. Let me compute 12*(0.242)^3 +14*(0.242)^2 -1 ≈ 12*(0.014) +14*(0.058) -1 ≈ 0.168 + 0.812 -1 ≈ 0.98 -1 ≈ -0.02. Wait, earlier we had that f(0.24) ≈ -0.0277 and f(0.245)≈0.0168. So the root is between 0.24 and 0.245. Let me check for the second cubic, 12x³ -14x² +1. Let me compute f(0.24) = 12*(0.24)^3 -14*(0.24)^2 +1 ≈ 12*(0.013824) -14*(0.0576) +1 ≈ 0.1659 - 0.8064 +1 ≈ 0.1659 + 0.1936 ≈ 0.3595. At x=0.24, f(x)= ~0.36. At x=0.25: f(0.25)=12*(0.015625) -14*(0.0625) +1 = 0.1875 - 0.875 +1 = 0.3125. At x=0.26: 12*(0.017576) -14*(0.0676) +1 ≈ 0.2109 - 0.9464 +1 ≈ 0.2645. At x=0.27: 12*(0.019683) -14*(0.0729) +1 ≈ 0.2362 - 1.0206 +1 ≈ 0.2156. At x=0.28: 12*(0.021952) -14*(0.0784) +1 ≈ 0.2634 - 1.0976 +1 ≈ 0.1658. At x=0.29: 12*(0.024389) -14*(0.0841) +1 ≈ 0.2927 - 1.1774 +1 ≈ 0.1153. At x=0.3: 12*(0.027) -14*(0.09) +1 = 0.324 - 1.26 +1 = 0.064. At x=0.31: 12*(0.029791) -14*(0.0961) +1 ≈ 0.3575 - 1.3454 +1 ≈ 0.0121. At x=0.315: 12*(0.03125) -14*(0.099225) +1 ≈ 0.375 - 1.389 +1 ≈ 0.375 - 0.389 ≈ -0.014. Therefore, between x=0.31 and x=0.315, f(x) crosses zero. Therefore, the second cubic has a root around 0.31, which is greater than 1/3 ≈ 0.333... Wait, 0.31 is less than 1/3 (≈0.333). Wait, 0.31 is less than 1/3. So the root is in (0.31, 0.315). But our original equation requires x ∈ (0, 1/3]. Therefore, this root is valid. However, we need to check which of the two cubics corresponds to the original equation. Let me check with the approximate value of x≈0.242. Let me plug into the second cubic: 12x³ -14x² +1. At x≈0.242:

12*(0.242)^3 -14*(0.242)^2 +1 ≈ 12*(0.014) -14*(0.058) +1 ≈ 0.168 - 0.812 +1 ≈ 0.356. Which is positive. Therefore, the root of the second cubic is at higher x, around 0.31, which is still within (0, 1/3]. Therefore, there are two solutions? But the original equation is arccos x + arccos 2x + arccos 3x = π. Let me check if both roots satisfy the original equation.

Let me take the root of the second cubic, say x≈0.31. Let me compute arccos(0.31) + arccos(0.62) + arccos(0.93). Let me approximate:

arccos(0.31) ≈ 1.266 radians (since cos(1.266) ≈ 0.31),

arccos(0.62) ≈ 0.916 radians (cos(0.916) ≈ 0.62),

arccos(0.93) ≈ 0.367 radians (cos(0.367) ≈ 0.93).

Sum ≈ 1.266 + 0.916 + 0.367 ≈ 2.549 radians, which is less than π ≈ 3.1416. Therefore, this does not satisfy the original equation. Therefore, this root is extraneous. Therefore, only the root of the first cubic, 12x³ +14x² -1 = 0, is valid. Therefore, the correct cubic equation is 12x³ +14x² -1 = 0. Therefore, the minimal polynomial is indeed this cubic, and 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|." Maybe there is another cubic equation with smaller coefficients? For example, if we can find a multiple of this equation with smaller coefficients? But since the coefficients are integers and the polynomial is irreducible, scaling would only increase the coefficients. For example, if we divide by a common factor, but there are no common factors. The coefficients are 12, 14, 0, -1. The GCD of 12, 14, 0, 1 is 1. Therefore, the polynomial is already in the simplest form. Therefore, the minimal sum is 27. However, the problem says "smallest possible value", so maybe there is another approach where we get a different cubic equation with smaller coefficients? Let me think again.

Alternatively, maybe I made a mistake in the initial steps? Let me check once more.

Original equation: arccos x + arccos 2x + arccos 3x = π. We set α + β + γ = π, then used α + β = π - γ, took cosine, and derived the equation leading to 12x³ +14x² -1 = 0. Then verified that this equation has a root in the required interval, and that the other root of the sixth-degree equation is extraneous. Therefore, the cubic equation is correct. Therefore, the answer should be 27. However, I recall that sometimes problems like this might have a different cubic equation if you consider different manipulations. Let me check if there's an alternative way.

Alternatively, suppose I let y = arccos x, then arccos 2x = π - y - arccos 3x. But not sure. Alternatively, use multiple angle formulas? Not sure. Alternatively, consider taking the original equation and using tan or something else. Let me try.

Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me take tan(α + β) = tan(π - γ) = -tan γ. But tan(α + β) = [tan α + tan β]/[1 - tan α tan β] = -tan γ. But tan α = sqrt(1 - x²)/x, since tan(arccos x) = sqrt(1 - x²)/x. Similarly, tan β = sqrt(1 - 4x²)/(2x), tan γ = sqrt(1 - 9x²)/(3x). Therefore:

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

Substituting:

[ 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 by 2x:

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

This seems even more complicated than before. Maybe cross-multiplying:

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

This seems very messy. Probably not helpful. Therefore, I think my initial approach was correct, leading to the cubic equation 12x³ +14x² -1 = 0. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, maybe multiplying by something or another approach.

Alternatively, maybe I can use substitution. Let me suppose that the cubic equation is 12x³ +14x² -1 = 0. Let me check if there is a different cubic equation with smaller coefficients. For example, if I can write it as something like 12x³ +14x² -1 = 0, but maybe there is a different monic polynomial? But since the coefficients are integers, and the polynomial is irreducible, the minimal monic polynomial would have larger coefficients. For example, if we divide by 12, we get x³ + (14/12)x² - 1/12 = 0, but coefficients are not integers. Therefore, not allowed. Therefore, the minimal polynomial with integer coefficients is indeed 12x³ +14x² -1. Therefore, the answer is 27.

But to be thorough, let me check if there is a cubic equation with smaller coefficients. For example, maybe coefficients with smaller absolute values. Let me think. Suppose someone claims that the minimal polynomial is 12x³ +14x² -1, but maybe there is another cubic equation with smaller coefficients. For example, if we can find a cubic equation with coefficients like 6x³ +7x² - something, but let me check.

Suppose we assume that the minimal polynomial is ax³ +bx² +cx +d = 0 with smaller coefficients. Let me think about possible coefficients. For example, if we can find a cubic equation with a=1, but then the equation would be x³ +bx² +cx +d = 0. But since we have a root in (0, 1/3], and the equation 12x³ +14x² -1 = 0, which is irreducible, it's unlikely that there is a monic cubic with integer coefficients. Similarly for other small a. For example, a=2: 2x³ +bx² +cx +d = 0. But unless the original equation can be scaled down, which it can't because of the coefficients. Therefore, I think 27 is the correct answer.

Therefore, the smallest possible value of |a| + |b| + |c| + |d| is 27.

**Final Answer**
\boxed{27}

To solve the equation:

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

we begin by analyzing the domain of the inverse cosine function. Since the range of $\arccos$ is $[0, \pi]$, and the arguments of the inverse cosine functions must lie in $[-1, 1]$, we deduce that:

- $x \in [-1, 1]$,
- $2x \in [-1, 1] \Rightarrow x \in [-0.5, 0.5]$,
- $3x \in [-1, 1] \Rightarrow x \in [-1/3, 1/3]$.

Thus, the only valid domain for $x$ is:

$$
x \in \left[ -\frac{1}{3}, \frac{1}{3} \right].
$$

However, since the sum of three arccosines equals $\pi$, and the arccos function is decreasing, we can infer that $x$ must be **positive**. If $x$ were negative, the sum of the arccosines would exceed $\pi$, which contradicts the given equation. Therefore, we restrict our attention to:

$$
x \in (0, \frac{1}{3}].
$$

---

### Step 1: Use Trigonometric Identities

Let:

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

Then the given equation becomes:

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

From this, we can write:

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

Taking the cosine of both sides:

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

Using the identity for $\cos(\alpha + \beta)$:

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

Substituting known values:

- $\cos \alpha = x$,
- $\cos \beta = 2x$,
- $\cos \gamma = 3x$,
- $\sin \alpha = \sqrt{1 - x^2}$,
- $\sin \beta = \sqrt{1 - 4x^2}$,
- $\sin \gamma = \sqrt{1 - 9x^2}$,

we get:

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

Simplifying:

$$
2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)} = -3x.
$$

Rearranging:

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

---

### Step 2: Square Both Sides

Squaring both sides:

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

Compute both sides:

- Left: $(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4$,
- Right: $(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2$.

Equating both sides:

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

Subtracting the left side from both sides:

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

Thus, the cubic equation is:

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

---

### Step 3: Verify the Polynomial

This cubic polynomial has integer coefficients and is irreducible over the rationals (as verified by the Rational Root Theorem and factoring attempts). Therefore, it is the minimal polynomial that $x$ satisfies.

---

### Step 4: Compute the Required Sum

We are asked to find the smallest possible value of:

$$
|a| + |b| + |c| + |d|,
$$

where the polynomial is of the form $ax^3 + bx^2 + cx + d = 0$.

From the polynomial $12x^3 + 14x^2 - 1 = 0$, we have:

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

So:

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

---

### Final Answer

$$
\boxed{27}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: arccos x + arccos 2x + arccos 3x = π. And then find the cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d|. Alright, let me start by recalling some trigonometric identities and properties of inverse cosine functions.

First, I know that the range of arccos is [0, π], so each term arccos x, arccos 2x, arccos 3x must be in that interval. Therefore, the arguments of arccos must be in [-1, 1]. So, for arccos x to be defined, x must be in [-1, 1]. Similarly, 2x must be in [-1, 1], so x ∈ [-0.5, 0.5]. Similarly, 3x must be in [-1, 1], so x ∈ [-1/3, 1/3]. Therefore, combining all these, x must be in [-1/3, 1/3]. So x is between -1/3 and 1/3. But since arccos is decreasing, and the sum of three arccos terms equals π, which is a positive number. Let me think about the possible values of x. If x is negative, then arccos x would be in (π/2, π), arccos 2x would be in (π/2, π) if 2x is negative, same with arccos 3x. But adding three angles each greater than π/2 would give more than 3π/2, which is more than π. So maybe x is positive? Let me check.

If x is positive, then arccos x, arccos 2x, arccos 3x are all in [0, π/2], since their arguments are positive and less than or equal to 1. Wait, but 3x is up to 1 when x is 1/3. So arccos(3x) would be arccos(1) = 0 when x = 1/3. So if x is positive, then arccos x is between 0 and π/2, arccos 2x is between 0 and π/2, arccos 3x is between 0 and π/2. So adding three angles between 0 and π/2. The maximum sum would be 3*(π/2) = 3π/2, but the equation says the sum is π. So maybe x is positive? Let me check if x can be negative. Suppose x is negative. Then arccos x is in (π/2, π), arccos 2x is in (π/2, π) if 2x is between -1 and 0, same with arccos 3x. So adding three angles each between π/2 and π would give a sum between 3π/2 and 3π. But the equation says the sum is π, which is less than 3π/2. Therefore, x cannot be negative. Therefore, x must be positive. Therefore, x ∈ (0, 1/3].

So x is between 0 and 1/3. So that's helpful. So we can focus on x in (0, 1/3].

Now, the equation is arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. So maybe I can use some trigonometric identities here. Let me think. If I take cos of both sides? Let me try that.

But cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) might be complicated. Alternatively, maybe I can use the identity for cos(α + β) = cos π - γ. Wait, since α + β + γ = π, then α + β = π - γ. Therefore, cos(α + β) = cos(π - γ) = -cos γ. So cos(α + β) = -cos γ. Let me compute cos(α + β) using the formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos(α + β) = -cos γ. Therefore:

cos α cos β - sin α sin β = -cos γ.

But cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x.

So 2x² - sin α sin β = -3x.

Therefore, sin α sin β = 2x² + 3x.

Hmm. Now, I need to find expressions for sin α and sin β. Since α = arccos x, then sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore:

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

So that's an equation in x. Let me square both sides to eliminate the square roots. But before that, let me note that squaring can introduce extraneous solutions, so I need to check later.

So squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2.

Let me compute both sides.

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

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

Therefore, equation becomes:

1 - 5x² + 4x^4 = 4x^4 + 12x^3 + 9x².

Subtracting left side from both sides:

0 = 4x^4 + 12x^3 + 9x² - (1 - 5x² + 4x^4) = 4x^4 + 12x^3 + 9x² -1 + 5x² -4x^4.

Simplify:

4x^4 -4x^4 cancels. Then 12x^3 + 9x² +5x² -1 = 12x^3 +14x² -1 = 0.

Therefore, the equation reduces to 12x³ +14x² -1 = 0. Wait, but hold on. Let me check my algebra again.

Original equation after squaring:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

So bringing left side to the right:

0 = 4x^4 +12x^3 +9x² -1 +5x² -4x^4.

Wait, no. Wait, original equation is left side equals right side. So:

Left side = right side => 1 -5x² +4x^4 = 4x^4 +12x^3 +9x².

Subtract left side from both sides:

0 = 4x^4 +12x^3 +9x² - (1 -5x² +4x^4) = 4x^4 +12x^3 +9x² -1 +5x² -4x^4.

Yes, so 4x^4 -4x^4 cancels. Then 12x^3 + (9x² +5x²) -1 = 12x^3 +14x² -1 = 0. Therefore, 12x³ +14x² -1 = 0. So that's a cubic equation. But wait, the problem says that x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0. So here, we have 12x³ +14x² +0x -1 = 0. Therefore, coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a cubic polynomial with smaller coefficients? Or maybe I made a mistake here?

Wait, but let me check if this equation is correct. Let me verify my steps again.

Starting from α + β + γ = π. Then α + β = π - γ. Then cos(α + β) = cos(π - γ) = -cos γ. Then expanding cos(α + β) = cos α cos β - sin α sin β = -cos γ. Then substituting cos α = x, cos β = 2x, cos γ = 3x. Therefore:

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

(1 - x²)(1 - 4x²) = (2x² + 3x)^2. Then expanding both sides:

Left side: 1 - 5x² + 4x^4.

Right side: 4x^4 + 12x^3 + 9x².

Subtracting left side from right side:

(4x^4 + 12x^3 + 9x²) - (1 -5x² +4x^4) = 12x^3 + 14x² -1 = 0. So that's correct. Therefore, the equation is 12x³ +14x² -1 = 0. Therefore, coefficients are 12, 14, 0, -1. So |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 is another way to get a different cubic equation with smaller coefficients? Or maybe I need to check if this equation is correct?

Wait, but maybe there are extraneous solutions introduced by squaring. So perhaps the original equation has solutions that don't satisfy the squared equation? Let me check.

So suppose x is a solution to 12x³ +14x² -1 = 0. Let me check if it satisfies the original equation. Let me try to find approximate solutions. Let me consider x in (0, 1/3]. Let me compute f(x) = 12x³ +14x² -1. At x=0, f(0) = -1. At x=1/3, f(1/3) = 12*(1/27) +14*(1/9) -1 = 12/27 +14/9 -1 = 4/9 +14/9 -1 = 18/9 -1 = 2 -1 = 1. Therefore, since f(0) = -1 and f(1/3) = 1, by Intermediate Value Theorem, there is a root between 0 and 1/3. Therefore, there is a solution. Let me check if that solution satisfies the original equation. But how?

Alternatively, maybe I can check if the equation 12x³ +14x² -1 = 0 is the minimal polynomial. But maybe there is a mistake in the derivation? Let me think again.

Alternatively, maybe I can use another approach. Let me try to use more identities. Let me consider that α + β + γ = π. So maybe I can take another pair of angles. For example, take α + γ = π - β. Then take cos(α + γ) = cos(π - β) = -cos β. So cos(α + γ) = -cos β. Let me compute cos(α + γ) = cos α cos γ - sin α sin γ. Which would be x*(3x) - sin α sin γ = 3x² - sin α sin γ = -cos β = -2x. Therefore:

3x² - sin α sin γ = -2x => sin α sin γ = 3x² + 2x.

Similarly, earlier we had sin α sin β = 2x² + 3x. So now we have two equations:

sin α sin β = 2x² + 3x,

sin α sin γ = 3x² + 2x.

Let me write them down:

Equation 1: sin α sin β = 2x² + 3x,

Equation 2: sin α sin γ = 3x² + 2x.

Also, we can do similar for β + γ = π - α. Let me try that. cos(β + γ) = cos(π - α) = -cos α = -x. Then cos(β + γ) = cos β cos γ - sin β sin γ = 2x * 3x - sin β sin γ = 6x² - sin β sin γ = -x. Therefore:

6x² - sin β sin γ = -x => sin β sin γ = 6x² + x.

So now we have three equations:

1. sin α sin β = 2x² + 3x,

2. sin α sin γ = 3x² + 2x,

3. sin β sin γ = 6x² + x.

So now, if I denote sin α = s, sin β = t, sin γ = u. Then:

s*t = 2x² + 3x,

s*u = 3x² + 2x,

t*u = 6x² + x.

So we have three equations with three variables s, t, u. Let me try to solve for s, t, u. Let me multiply all three equations together:

(s*t)*(s*u)*(t*u) = (2x² + 3x)(3x² + 2x)(6x² + x).

Left side: s² t² u² = (s t u)^2.

Right side: Let me compute that. Let me first compute (2x² + 3x)(3x² + 2x). Let me compute that first:

(2x² + 3x)(3x² + 2x) = 2x²*3x² + 2x²*2x + 3x*3x² + 3x*2x = 6x^4 + 4x³ + 9x³ + 6x² = 6x^4 + 13x³ + 6x².

Then multiply by (6x² + x):

(6x^4 + 13x³ + 6x²)(6x² + x) = Let me compute term by term:

First, 6x^4 * 6x² = 36x^6,

6x^4 * x = 6x^5,

13x³ * 6x² = 78x^5,

13x³ * x = 13x^4,

6x² * 6x² = 36x^4,

6x² * x = 6x³.

Adding all together:

36x^6 + (6x^5 + 78x^5) + (13x^4 + 36x^4) + 6x³ = 36x^6 + 84x^5 + 49x^4 + 6x³.

Therefore, the right side is 36x^6 + 84x^5 + 49x^4 + 6x³.

On the left side, (s t u)^2. Let me find s t u. Let me note that from equations:

From equation 1: s t = 2x² + 3x,

From equation 2: s u = 3x² + 2x,

From equation 3: t u = 6x² + x.

Let me solve for s, t, u. Let me take the product of equations 1, 2, 3:

(s t)(s u)(t u) = (2x² + 3x)(3x² + 2x)(6x² + x) = (s t u)^2, which we already calculated as 36x^6 + 84x^5 + 49x^4 + 6x³. Therefore, s t u = sqrt(36x^6 + 84x^5 + 49x^4 + 6x³). But maybe we can find s t u another way?

Alternatively, let me solve for s, t, u. Let me take equation 1: s t = 2x² + 3x,

equation 2: s u = 3x² + 2x,

equation 3: t u = 6x² + x.

Let me solve for s, t, u. Let me divide equation 2 by equation 1: (s u)/(s t) = (3x² + 2x)/(2x² + 3x) => u/t = (3x² + 2x)/(2x² + 3x). Therefore, u = t*(3x² + 2x)/(2x² + 3x). Similarly, from equation 3: t u = 6x² + x. Substitute u:

t * [ t*(3x² + 2x)/(2x² + 3x) ] = 6x² + x => t² * (3x² + 2x)/(2x² + 3x) = 6x² + x.

Therefore, t² = (6x² + x) * (2x² + 3x)/(3x² + 2x). Let me compute that:

First, note that 6x² + x = x(6x + 1), and 2x² + 3x = x(2x + 3), and 3x² + 2x = x(3x + 2). Therefore:

t² = [x(6x + 1) * x(2x + 3)] / [x(3x + 2)] = [x²(6x + 1)(2x + 3)] / [x(3x + 2)] = x(6x + 1)(2x + 3)/(3x + 2).

Therefore, t² = x(6x + 1)(2x + 3)/(3x + 2). Similarly, since t = sin β, which is sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore, t² = 1 - 4x². Therefore:

1 - 4x² = x(6x + 1)(2x + 3)/(3x + 2).

Let me compute the right side:

First compute (6x + 1)(2x + 3) = 12x² + 18x + 2x + 3 = 12x² + 20x + 3.

Therefore, right side is x*(12x² + 20x + 3)/(3x + 2). Therefore:

1 - 4x² = [x(12x² + 20x + 3)] / (3x + 2).

Multiply both sides by (3x + 2):

(1 - 4x²)(3x + 2) = x(12x² + 20x + 3).

Let me expand left side:

1*(3x + 2) - 4x²*(3x + 2) = 3x + 2 - 12x³ - 8x².

Right side: 12x³ + 20x² + 3x.

Therefore, equation becomes:

3x + 2 - 12x³ - 8x² = 12x³ + 20x² + 3x.

Subtract left side from both sides:

0 = 12x³ + 20x² + 3x - (3x + 2 - 12x³ - 8x²) = 12x³ + 20x² + 3x -3x -2 +12x³ +8x².

Simplify:

12x³ +12x³ = 24x³,

20x² +8x² = 28x²,

3x -3x = 0,

-2.

Therefore, 24x³ +28x² -2 = 0. Divide both sides by 2: 12x³ +14x² -1 = 0. Which is the same equation as before! So that confirms that the equation is correct. Therefore, the cubic equation is indeed 12x³ +14x² -1 = 0. Therefore, the coefficients are a=12, b=14, c=0, d=-1. Therefore, |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a different cubic equation with smaller coefficients? Or maybe this is the minimal one?

Wait, but maybe the equation can be scaled? For example, if we multiply by some integer, but the problem says "a cubic polynomial of the form ax³ +bx² +cx +d = 0, where a, b, c, and d are integers, and a ≠ 0." So the minimal polynomial is 12x³ +14x² -1, but maybe there is another polynomial with smaller coefficients? For example, if there is a common factor? Let me check if 12x³ +14x² -1 can be factored. Let me try rational roots. By Rational Root Theorem, possible rational roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12. Let me test them.

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*(-1/2)^3 +14*(-1/2)^2 -1 = 12*(-1/8) +14*(1/4) -1 = -1.5 + 3.5 -1 = 1 ≠0.

x=1/3: 12*(1/27) +14*(1/9) -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 = -4/9 +14/9 -1 = 10/9 -1 = 1/9 ≠0.

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 ≠0.

x=-1/4: 12*(-1/4)^3 +14*(-1/4)^2 -1 = 12*(-1/64) +14*(1/16) -1 = -3/16 +7/8 -1 = -3/16 +14/16 -16/16 = (-3 +14 -16)/16 = (-5)/16 ≠0.

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 ≠0.

x=-1/6: 12*(-1/6)^3 +14*(-1/6)^2 -1 = 12*(-1/216) +14*(1/36) -1 = -1/18 +7/18 -1 = 6/18 -1 = 1/3 -1 = -2/3 ≠0.

x=1/12: 12*(1/12)^3 +14*(1/12)^2 -1 = 12*(1/1728) +14*(1/144) -1 = 1/144 + 7/72 -1 = 1/144 + 14/144 -144/144 = (1 +14 -144)/144 = (-129)/144 ≠0.

Similarly for others. Therefore, there are no rational roots. Therefore, the polynomial 12x³ +14x² -1 is irreducible over rationals? Therefore, it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial. Therefore, the coefficients are 12, 14, 0, -1, so the sum is 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." Maybe there is another approach where we get a different cubic equation with smaller coefficients? Let me think.

Alternatively, maybe I made a mistake in assuming that squaring didn't introduce extraneous solutions? Let me check if the equation 12x³ +14x² -1 = 0 has solutions in (0, 1/3]. Let me approximate the root. Let me take f(x) = 12x³ +14x² -1. At x=0.2: f(0.2)=12*(0.008) +14*(0.04) -1 = 0.096 + 0.56 -1 = -0.344. At x=0.25: f(0.25)=12*(0.015625) +14*(0.0625) -1 = 0.1875 + 0.875 -1 = 1.0625 -1 = 0.0625. So between 0.2 and 0.25, f(x) crosses zero. Let me try x=0.23: f(0.23)=12*(0.012167) +14*(0.0529) -1 ≈ 0.146 + 0.7406 -1 ≈ 0.8866 -1 ≈ -0.1134. x=0.24: 12*(0.013824) +14*(0.0576) -1 ≈ 0.1659 + 0.8064 -1 ≈ 0.9723 -1 ≈ -0.0277. x=0.245: 12*(0.245)^3 +14*(0.245)^2 -1. Let me compute 0.245^2 = 0.060025, 0.245^3 ≈ 0.014706. Then 12*0.014706 ≈ 0.1765, 14*0.060025 ≈ 0.84035. Total ≈ 0.1765 + 0.84035 -1 ≈ 1.01685 -1 ≈ 0.01685. So between 0.24 and 0.245, f(x) crosses zero. So the root is approximately 0.242. Let me check if this value satisfies the original equation arccos x + arccos 2x + arccos 3x = π. Let me compute approximate values.

Let me take x ≈ 0.242. Then:

arccos(0.242) ≈ let me compute. Since cos(π/3) = 0.5, cos(π/2) = 0. So arccos(0.242) is between π/3 ≈ 1.047 and π/2 ≈ 1.5708. Let me use calculator-like approximations. Let me recall that cos(1.35 radians) ≈ cos(77.3 degrees) ≈ 0.2225. Wait, 0.242 is larger than 0.2225, so arccos(0.242) is less than 1.35 radians. Let me try 1.3 radians: cos(1.3) ≈ 0.2675. 1.3 radians is about 74.5 degrees. So cos(1.3) ≈ 0.2675. Our x is 0.242, which is less than 0.2675, so arccos(0.242) is greater than 1.3 radians. Let me try 1.33 radians: cos(1.33) ≈ cos(76.2 degrees). Let me compute cos(1.33):

Using Taylor series or calculator? Well, since I don't have a calculator here, but let me approximate. Let me use linear approximation between 1.3 and 1.35. At 1.3, cos(1.3) ≈ 0.2675. At 1.35, cos(1.35) ≈ 0.2225. The difference between 1.3 and 1.35 is 0.05 radians, and the difference in cos is 0.2225 - 0.2675 = -0.045. So slope is -0.045 / 0.05 = -0.9 per radian. So to find x where cos(x) = 0.242, starting from 1.3 where cos(1.3)=0.2675. Let me set up linear approximation:

Let me denote f(x) = cos(x). We have f(1.3) = 0.2675, f'(x) = -sin(x). At x=1.3, sin(1.3) ≈ sin(74.5 degrees) ≈ 0.9613. Therefore, f'(1.3) ≈ -0.9613. So linear approximation:

f(x) ≈ f(1.3) + f'(1.3)(x - 1.3). We want f(x) = 0.242. Therefore:

0.242 ≈ 0.2675 - 0.9613*(x - 1.3)

=> 0.242 - 0.2675 ≈ -0.9613*(x - 1.3)

=> -0.0255 ≈ -0.9613*(x - 1.3)

=> x - 1.3 ≈ 0.0255 / 0.9613 ≈ 0.0265

=> x ≈ 1.3 + 0.0265 ≈ 1.3265 radians.

So arccos(0.242) ≈ 1.3265 radians.

Similarly, arccos(2x) = arccos(0.484). Let me compute that. cos(π/3) = 0.5, so arccos(0.484) is slightly larger than π/3 ≈ 1.047 radians. Let me approximate. Let me take x=1.047 radians, cos(1.047)=0.5. Let me try 1.05 radians: cos(1.05) ≈ cos(60.2 degrees). Let me compute cos(1.05):

Again, using linear approximation. Let me take x=1.047 (π/3), cos(π/3)=0.5. Let me take derivative at π/3: -sin(π/3) ≈ -√3/2 ≈ -0.866. So f(x) = cos(x). Let me approximate cos(1.05):

Let me write 1.05 = π/3 + Δx, where Δx = 1.05 - π/3 ≈ 1.05 - 1.047 ≈ 0.003 radians. Then cos(π/3 + Δx) ≈ cos(π/3) - sin(π/3)*Δx ≈ 0.5 - 0.866*0.003 ≈ 0.5 - 0.0026 ≈ 0.4974. But we need cos(x) = 0.484, which is less than 0.4974. Therefore, need to go further. Let me try Δx = 0.01 radians. Then cos(π/3 + 0.01) ≈ 0.5 - 0.866*0.01 ≈ 0.5 - 0.00866 ≈ 0.4913. Still higher than 0.484. Let me try Δx = 0.015 radians: cos(π/3 + 0.015) ≈ 0.5 - 0.866*0.015 ≈ 0.5 - 0.01299 ≈ 0.487. Closer. Need 0.484. Let me try Δx = 0.017 radians: 0.5 - 0.866*0.017 ≈ 0.5 - 0.0147 ≈ 0.4853. Still higher. Δx=0.018: 0.5 - 0.866*0.018 ≈ 0.5 - 0.0156 ≈ 0.4844. Close. So arccos(0.484) ≈ π/3 + 0.018 ≈ 1.047 + 0.018 ≈ 1.065 radians.

Similarly, arccos(3x) = arccos(0.726). Let me compute that. cos(π/4) ≈ 0.707, so arccos(0.726) is slightly less than π/4 ≈ 0.785 radians. Let me approximate. Let me take x=π/4 ≈ 0.785, cos(π/4)=√2/2 ≈ 0.707. Let me compute cos(0.75 radians): cos(0.75) ≈ 0.7317. Which is higher than 0.726. Let me try 0.76 radians: cos(0.76) ≈ cos(43.5 degrees). Let me use linear approximation. Let me take x=0.75, cos(0.75)=0.7317. Let me take derivative at x=0.75: -sin(0.75) ≈ -sin(43 degrees) ≈ -0.682. So f(x) = cos(x). Let me approximate cos(0.75 + Δx) ≈ 0.7317 - 0.682*Δx. We want cos(x) = 0.726. So:

0.726 ≈ 0.7317 - 0.682*Δx => Δx ≈ (0.7317 - 0.726)/0.682 ≈ 0.0057 / 0.682 ≈ 0.00836 radians. Therefore, arccos(0.726) ≈ 0.75 + 0.00836 ≈ 0.7584 radians.

Therefore, sum of arccos x + arccos 2x + arccos 3x ≈ 1.3265 + 1.065 + 0.7584 ≈ 1.3265 + 1.065 = 2.3915 + 0.7584 ≈ 3.1499 radians. Which is approximately π ≈ 3.1416. So the approximate value is about 3.15, which is slightly larger than π. Therefore, there might be some error due to my approximations. But it's close. Therefore, the root of the equation 12x³ +14x² -1 = 0 is indeed a solution to the original equation. Therefore, the cubic equation is correct.

But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is another cubic equation with smaller coefficients? For example, if we can factor out something? But since the polynomial is irreducible over rationals, as we saw, there are no rational roots, so it cannot be factored into a product of lower-degree polynomials with integer coefficients. Therefore, the minimal polynomial is indeed 12x³ +14x² -1. Therefore, the sum is 12 +14 +0 +1 = 27. However, maybe there is another approach where we get a different cubic equation? Let me think again.

Alternatively, maybe I can use another identity. Let me consider taking the original equation arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me take the first two terms: α + β = π - γ. Then take cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ. Which is what I did before. So that gives me the equation leading to 12x³ +14x² -1 = 0. Alternatively, maybe I can use sine? Let me think. Since α + β + γ = π, then maybe take sine of both sides? Let me try:

sin(α + β + γ) = sin π = 0. But expanding sin(α + β + γ) is complicated. Alternatively, since α + β = π - γ, then sin(α + β) = sin(π - γ) = sin γ. So sin(α + β) = sin γ. Expanding sin(α + β) = sin α cos β + cos α sin β = sin γ. Therefore:

sin α cos β + cos α sin β = sin γ.

We know cos α = x, cos β = 2x, cos γ = 3x. Also, sin α = sqrt(1 - x²), sin β = sqrt(1 - 4x²), sin γ = sqrt(1 - 9x²). Therefore:

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

This seems another equation. Let me write it:

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

This seems more complicated, but maybe squaring both sides? Let me try. Let me denote LHS = 2x sqrt(1 - x²) + x sqrt(1 - 4x²), RHS = sqrt(1 - 9x²). Squaring both sides:

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

Expanding the left side:

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

Compute each term:

First term: 4x²(1 - x²).

Second term: 4x² sqrt{(1 - x²)(1 - 4x²)}.

Third term: x²(1 - 4x²).

Therefore, total left side:

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

Simplify the first two terms:

4x² - 4x^4 + x² - 4x^4 = 5x² - 8x^4.

Therefore, equation becomes:

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

Bring the non-radical terms to the right:

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

Therefore:

sqrt{(1 - x²)(1 - 4x²)} = [1 -14x² +8x^4]/(4x²).

Square both sides again:

(1 - x²)(1 - 4x²) = [ (1 -14x² +8x^4)^2 ] / (16x^4).

Multiply both sides by 16x^4:

16x^4(1 - x²)(1 - 4x²) = (1 -14x² +8x^4)^2.

This seems very complicated, but let me compute both sides.

Left side: 16x^4*(1 - x²)(1 - 4x²). Earlier, we had (1 - x²)(1 - 4x²) = 1 -5x² +4x^4. Therefore, left side is 16x^4*(1 -5x² +4x^4) = 16x^4 -80x^6 +64x^8.

Right side: (1 -14x² +8x^4)^2. Let me expand this:

Let me denote A = 8x^4 -14x² +1. Then A² = (8x^4)^2 + (-14x²)^2 +1^2 + 2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1).

Wait, more systematically:

(8x^4 -14x² +1)^2 = (8x^4)^2 + (-14x²)^2 + (1)^2 + 2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1).

Compute term by term:

(8x^4)^2 = 64x^8,

(-14x²)^2 = 196x^4,

(1)^2 = 1,

Cross terms:

2*(8x^4*(-14x²) +8x^4*1 + (-14x²)*1) = 2*(-112x^6 +8x^4 -14x²) = -224x^6 +16x^4 -28x².

Therefore, total right side:

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

Therefore, equation is:

Left side: 16x^4 -80x^6 +64x^8 = Right side: 64x^8 -224x^6 +212x^4 -28x² +1.

Subtract left side from both sides:

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

Simplify:

64x^8 -64x^8 = 0,

-224x^6 +80x^6 = -144x^6,

212x^4 -16x^4 = 196x^4,

-28x²,

+1.

Therefore, equation becomes:

-144x^6 +196x^4 -28x² +1 = 0.

Multiply both sides by -1:

144x^6 -196x^4 +28x² -1 = 0.

This is a sixth-degree equation. However, we already have a cubic equation from before. Let me check if this sixth-degree equation is related to the cubic equation. Let me note that if x satisfies 12x³ +14x² -1 = 0, then maybe substituting x³ = (-14x² +1)/12 into this equation. Let me try.

Let me denote the sixth-degree equation: 144x^6 -196x^4 +28x² -1 = 0. Let me express x^6 in terms of lower degrees. From the cubic equation 12x³ +14x² -1 = 0, we can solve for x³:

x³ = (1 -14x²)/12.

Therefore, x^6 = (x³)^2 = [(1 -14x²)/12]^2 = (1 -28x² +196x^4)/144.

Substitute into the sixth-degree equation:

144x^6 -196x^4 +28x² -1 = 144*(1 -28x² +196x^4)/144 -196x^4 +28x² -1 = (1 -28x² +196x^4) -196x^4 +28x² -1 = 1 -28x² +196x^4 -196x^4 +28x² -1 = 0. Therefore, the sixth-degree equation is satisfied by the roots of the cubic equation. Therefore, the sixth-degree equation is a multiple of the cubic equation? Or maybe the cubic equation is a factor? Let me check.

Let me perform polynomial division: divide 144x^6 -196x^4 +28x² -1 by 12x³ +14x² -1. Let me see.

Let me write the sixth-degree polynomial as 144x^6 + 0x^5 -196x^4 +0x^3 +28x² +0x -1.

Divide by 12x³ +14x² -1.

Let me use polynomial long division.

Divide 144x^6 by 12x³: 12x³. Multiply divisor by 12x³: 12x³*(12x³ +14x² -1) = 144x^6 +168x^5 -12x³.

Subtract this from the dividend:

(144x^6 +0x^5 -196x^4 +0x^3 +28x² +0x -1) - (144x^6 +168x^5 -12x³) = 0x^6 -168x^5 -196x^4 +12x³ +28x² +0x -1.

Next term: -168x^5 divided by 12x³ is -14x². Multiply divisor by -14x²: -14x²*(12x³ +14x² -1) = -168x^5 -196x^4 +14x².

Subtract this from the current dividend:

(-168x^5 -196x^4 +12x³ +28x² +0x -1) - (-168x^5 -196x^4 +14x²) = 0x^5 +0x^4 +12x³ + (28x² -14x²) +0x -1 = 12x³ +14x² -1.

Now, divide 12x³ +14x² -1 by the divisor 12x³ +14x² -1, which is 1. Therefore, the division yields quotient 12x³ -14x² +1, and remainder zero. Therefore, the sixth-degree polynomial factors as (12x³ +14x² -1)(12x³ -14x² +1). Therefore, the sixth-degree equation factors into two cubics. Therefore, the original equation 144x^6 -196x^4 +28x² -1 = 0 is equivalent to (12x³ +14x² -1)(12x³ -14x² +1) = 0. Therefore, the solutions are the roots of 12x³ +14x² -1 = 0 and 12x³ -14x² +1 = 0. However, we need to check which of these cubics have roots in the interval (0, 1/3].

Let me check the second cubic: 12x³ -14x² +1 = 0. Let me check for roots in (0, 1/3]. Let me define f(x) = 12x³ -14x² +1. At x=0: f(0)=1. At x=1/3: f(1/3)=12*(1/27) -14*(1/9) +1 = 4/9 -14/9 +1 = (-10/9) +1 = -1/9. Therefore, since f(0)=1 and f(1/3)=-1/9, there is a root between 0 and 1/3. Therefore, both cubics have roots in (0, 1/3]. However, we need to check which one is the correct one. Let me check with the approximate value we had earlier, x≈0.242. Let me plug into 12x³ +14x² -1 ≈ 0. Let me compute 12*(0.242)^3 +14*(0.242)^2 -1 ≈ 12*(0.014) +14*(0.058) -1 ≈ 0.168 + 0.812 -1 ≈ 0.98 -1 ≈ -0.02. Wait, earlier we had that f(0.24) ≈ -0.0277 and f(0.245)≈0.0168. So the root is between 0.24 and 0.245. Let me check for the second cubic, 12x³ -14x² +1. Let me compute f(0.24) = 12*(0.24)^3 -14*(0.24)^2 +1 ≈ 12*(0.013824) -14*(0.0576) +1 ≈ 0.1659 - 0.8064 +1 ≈ 0.1659 + 0.1936 ≈ 0.3595. At x=0.24, f(x)= ~0.36. At x=0.25: f(0.25)=12*(0.015625) -14*(0.0625) +1 = 0.1875 - 0.875 +1 = 0.3125. At x=0.26: 12*(0.017576) -14*(0.0676) +1 ≈ 0.2109 - 0.9464 +1 ≈ 0.2645. At x=0.27: 12*(0.019683) -14*(0.0729) +1 ≈ 0.2362 - 1.0206 +1 ≈ 0.2156. At x=0.28: 12*(0.021952) -14*(0.0784) +1 ≈ 0.2634 - 1.0976 +1 ≈ 0.1658. At x=0.29: 12*(0.024389) -14*(0.0841) +1 ≈ 0.2927 - 1.1774 +1 ≈ 0.1153. At x=0.3: 12*(0.027) -14*(0.09) +1 = 0.324 - 1.26 +1 = 0.064. At x=0.31: 12*(0.029791) -14*(0.0961) +1 ≈ 0.3575 - 1.3454 +1 ≈ 0.0121. At x=0.315: 12*(0.03125) -14*(0.099225) +1 ≈ 0.375 - 1.389 +1 ≈ 0.375 - 0.389 ≈ -0.014. Therefore, between x=0.31 and x=0.315, f(x) crosses zero. Therefore, the second cubic has a root around 0.31, which is greater than 1/3 ≈ 0.333... Wait, 0.31 is less than 1/3 (≈0.333). Wait, 0.31 is less than 1/3. So the root is in (0.31, 0.315). But our original equation requires x ∈ (0, 1/3]. Therefore, this root is valid. However, we need to check which of the two cubics corresponds to the original equation. Let me check with the approximate value of x≈0.242. Let me plug into the second cubic: 12x³ -14x² +1. At x≈0.242:

12*(0.242)^3 -14*(0.242)^2 +1 ≈ 12*(0.014) -14*(0.058) +1 ≈ 0.168 - 0.812 +1 ≈ 0.356. Which is positive. Therefore, the root of the second cubic is at higher x, around 0.31, which is still within (0, 1/3]. Therefore, there are two solutions? But the original equation is arccos x + arccos 2x + arccos 3x = π. Let me check if both roots satisfy the original equation.

Let me take the root of the second cubic, say x≈0.31. Let me compute arccos(0.31) + arccos(0.62) + arccos(0.93). Let me approximate:

arccos(0.31) ≈ 1.266 radians (since cos(1.266) ≈ 0.31),

arccos(0.62) ≈ 0.916 radians (cos(0.916) ≈ 0.62),

arccos(0.93) ≈ 0.367 radians (cos(0.367) ≈ 0.93).

Sum ≈ 1.266 + 0.916 + 0.367 ≈ 2.549 radians, which is less than π ≈ 3.1416. Therefore, this does not satisfy the original equation. Therefore, this root is extraneous. Therefore, only the root of the first cubic, 12x³ +14x² -1 = 0, is valid. Therefore, the correct cubic equation is 12x³ +14x² -1 = 0. Therefore, the minimal polynomial is indeed this cubic, and 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|." Maybe there is another cubic equation with smaller coefficients? For example, if we can find a multiple of this equation with smaller coefficients? But since the coefficients are integers and the polynomial is irreducible, scaling would only increase the coefficients. For example, if we divide by a common factor, but there are no common factors. The coefficients are 12, 14, 0, -1. The GCD of 12, 14, 0, 1 is 1. Therefore, the polynomial is already in the simplest form. Therefore, the minimal sum is 27. However, the problem says "smallest possible value", so maybe there is another approach where we get a different cubic equation with smaller coefficients? Let me think again.

Alternatively, maybe I made a mistake in the initial steps? Let me check once more.

Original equation: arccos x + arccos 2x + arccos 3x = π. We set α + β + γ = π, then used α + β = π - γ, took cosine, and derived the equation leading to 12x³ +14x² -1 = 0. Then verified that this equation has a root in the required interval, and that the other root of the sixth-degree equation is extraneous. Therefore, the cubic equation is correct. Therefore, the answer should be 27. However, I recall that sometimes problems like this might have a different cubic equation if you consider different manipulations. Let me check if there's an alternative way.

Alternatively, suppose I let y = arccos x, then arccos 2x = π - y - arccos 3x. But not sure. Alternatively, use multiple angle formulas? Not sure. Alternatively, consider taking the original equation and using tan or something else. Let me try.

Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me take tan(α + β) = tan(π - γ) = -tan γ. But tan(α + β) = [tan α + tan β]/[1 - tan α tan β] = -tan γ. But tan α = sqrt(1 - x²)/x, since tan(arccos x) = sqrt(1 - x²)/x. Similarly, tan β = sqrt(1 - 4x²)/(2x), tan γ = sqrt(1 - 9x²)/(3x). Therefore:

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

Substituting:

[ 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 by 2x:

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

This seems even more complicated than before. Maybe cross-multiplying:

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

This seems very messy. Probably not helpful. Therefore, I think my initial approach was correct, leading to the cubic equation 12x³ +14x² -1 = 0. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, maybe multiplying by something or another approach.

Alternatively, maybe I can use substitution. Let me suppose that the cubic equation is 12x³ +14x² -1 = 0. Let me check if there is a different cubic equation with smaller coefficients. For example, if I can write it as something like 12x³ +14x² -1 = 0, but maybe there is a different monic polynomial? But since the coefficients are integers, and the polynomial is irreducible, the minimal monic polynomial would have larger coefficients. For example, if we divide by 12, we get x³ + (14/12)x² - 1/12 = 0, but coefficients are not integers. Therefore, not allowed. Therefore, the minimal polynomial with integer coefficients is indeed 12x³ +14x² -1. Therefore, the answer is 27.

But to be thorough, let me check if there is a cubic equation with smaller coefficients. For example, maybe coefficients with smaller absolute values. Let me think. Suppose someone claims that the minimal polynomial is 12x³ +14x² -1, but maybe there is another cubic equation with smaller coefficients. For example, if we can find a cubic equation with coefficients like 6x³ +7x² - something, but let me check.

Suppose we assume that the minimal polynomial is ax³ +bx² +cx +d = 0 with smaller coefficients. Let me think about possible coefficients. For example, if we can find a cubic equation with a=1, but then the equation would be x³ +bx² +cx +d = 0. But since we have a root in (0, 1/3], and the equation 12x³ +14x² -1 = 0, which is irreducible, it's unlikely that there is a monic cubic with integer coefficients. Similarly for other small a. For example, a=2: 2x³ +bx² +cx +d = 0. But unless the original equation can be scaled down, which it can't because of the coefficients. Therefore, I think 27 is the correct answer.

Therefore, the smallest possible value of |a| + |b| + |c| + |d| is 27.

**Final Answer**
\boxed{27}

To solve the equation:

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

we begin by analyzing the domain of the inverse cosine function. Since the range of $\arccos$ is $[0, \pi]$, and the arguments of the inverse cosine functions must lie in $[-1, 1]$, we deduce that:

- $x \in [-1, 1]$,
- $2x \in [-1, 1] \Rightarrow x \in [-0.5, 0.5]$,
- $3x \in [-1, 1] \Rightarrow x \in [-1/3, 1/3]$.

Thus, the only valid domain for $x$ is:

$$
x \in \left[ -\frac{1}{3}, \frac{1}{3} \right].
$$

However, since the sum of three arccosines equals $\pi$, and the arccos function is decreasing, we can infer that $x$ must be **positive**. If $x$ were negative, the sum of the arccosines would exceed $\pi$, which contradicts the given equation. Therefore, we restrict our attention to:

$$
x \in (0, \frac{1}{3}].
$$

---

### Step 1: Use Trigonometric Identities

Let:

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

Then the given equation becomes:

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

From this, we can write:

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

Taking the cosine of both sides:

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

Using the identity for $\cos(\alpha + \beta)$:

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

Substituting known values:

- $\cos \alpha = x$,
- $\cos \beta = 2x$,
- $\cos \gamma = 3x$,
- $\sin \alpha = \sqrt{1 - x^2}$,
- $\sin \beta = \sqrt{1 - 4x^2}$,
- $\sin \gamma = \sqrt{1 - 9x^2}$,

we get:

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

Simplifying:

$$
2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)} = -3x.
$$

Rearranging:

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

---

### Step 2: Square Both Sides

Squaring both sides:

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

Compute both sides:

- Left: $(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4$,
- Right: $(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2$.

Equating both sides:

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

Subtracting the left side from both sides:

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

Thus, the cubic equation is:

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

---

### Step 3: Verify the Polynomial

This cubic polynomial has integer coefficients and is irreducible over the rationals (as verified by the Rational Root Theorem and factoring attempts). Therefore, it is the minimal polynomial that $x$ satisfies.

---

### Step 4: Compute the Required Sum

We are asked to find the smallest possible value of:

$$
|a| + |b| + |c| + |d|,
$$

where the polynomial is of the form $ax^3 + bx^2 + cx + d = 0$.

From the polynomial $12x^3 + 14x^2 - 1 = 0$, we have:

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

So:

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

---

### 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 $12 + 14 + 0 + 1 = \boxed{27}.$