Let `cos^(-1)(x)+cos^(-1)(2x)+cos^(-1)(3x)b epidot` If `x` satisfies the equation `a x^3+b x^2+c x-c_1=0,` then the value of `(b-a-c)` is_________

Explanation

To solve the problem, we need to analyze the equation given and manipulate it step by step.

### Step 1: Write down the equation
We start with the equation:
\[ \cos^{-1}(x) + \cos^{-1}(2x) + \cos^{-1}(3x) = \pi \]

### Step 2: Rearrange the equation
From the equation, we can isolate one of the inverse cosine terms:
\[ \cos^{-1}(2x) + \cos^{-1}(3x) = \pi - \cos^{-1}(x) \]
Using the property of inverse cosine, we can rewrite \(\pi - \cos^{-1}(x)\) as \(\cos^{-1}(-x)\):
\[ \cos^{-1}(2x) + \cos^{-1}(3x) = \cos^{-1}(-x) \]

### Step 3: Apply the formula for the sum of inverse cosines
Using the formula for the sum of inverse cosines:
\[ \cos^{-1}(a) + \cos^{-1}(b) = \cos^{-1}(ab - \sqrt{(1-a^2)(1-b^2)}) \]
we set \(a = 2x\) and \(b = 3x\):
\[ \cos^{-1}(2x) + \cos^{-1}(3x) = \cos^{-1}(2x \cdot 3x - \sqrt{(1 - (2x)^2)(1 - (3x)^2)}) \]
This simplifies to:
\[ \cos^{-1}(6x^2 - \sqrt{(1 - 4x^2)(1 - 9x^2)}) \]

### Step 4: Set the equation for the cosine
Now we set:
\[ 6x^2 - \sqrt{(1 - 4x^2)(1 - 9x^2)} = -x \]
Rearranging gives:
\[ \sqrt{(1 - 4x^2)(1 - 9x^2)} = 6x^2 + x \]

### Step 5: Square both sides
Squaring both sides results in:
\[ (1 - 4x^2)(1 - 9x^2) = (6x^2 + x)^2 \]

### Step 6: Expand both sides
Expanding the left side:
\[ 1 - 9x^2 - 4x^2 + 36x^4 = 1 - 13x^2 + 36x^4 \]
And the right side:
\[ (6x^2 + x)^2 = 36x^4 + 12x^3 + x^2 \]

### Step 7: Set the equation to zero
Equating both sides gives:
\[ 1 - 13x^2 + 36x^4 = 36x^4 + 12x^3 + x^2 \]
This simplifies to:
\[ -13x^2 - 12x^3 + 1 = 0 \]
Rearranging gives:
\[ 12x^3 + 14x^2 - 1 = 0 \]

### Step 8: Identify coefficients
From the equation \(12x^3 + 14x^2 + 0x - 1 = 0\), we identify:
- \(a = 12\)
- \(b = 14\)
- \(c = 0\)

### Step 9: Calculate \(b - a - c\)
Now we compute:
\[ b - a - c = 14 - 12 - 0 = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]