To solve the equation
\[ \frac{x - 6}{x^3 + 2x} = \frac{A}{x} + \frac{B}{x^2} + C + \frac{2}{x^2}, \]
we first need to simplify the left-hand side. We can factor the denominator \(x^3 + 2x\):
\[ x^3 + 2x = x(x^2 + 2). \]
Hence, we rewrite the expression:
\[ \frac{x - 6}{x(x^2 + 2)}. \]
Now, we want to express the right-hand side in a common denominator of \(x(x^2 + 2)\), so we rewrite each term accordingly:
\[ \frac{A}{x} = \frac{A(x^2 + 2)}{x(x^2 + 2)}, \] \[ \frac{B}{x^2} = \frac{B(x)}{x(x^2 + 2)}, \] \[ C = \frac{C \cdot x(x^2 + 2)}{x(x^2 + 2)}, \] \[ \frac{2}{x^2} = \frac{2(x)}{x(x^2 + 2)}. \]
Combining all these, the right-hand side becomes:
\[ \frac{A(x^2 + 2) + Bx + Cx(x^2 + 2) + 2x}{x(x^2 + 2)}. \]
Next, equate the numerators from both sides:
\[ x - 6 = A(x^2 + 2) + Bx + Cx(x^2 + 2) + 2x. \]
Now we need to expand and simplify:
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Expand \( A(x^2 + 2) \): \[ Ax^2 + 2A. \]
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The term \( Cx(x^2 + 2) \) becomes: \[ Cx^3 + 2Cx. \]
Now substituting back, our equation becomes:
\[ x - 6 = (C)x^3 + (A + 2C + B)x^2 + (2A + 2C)x. \]
Setting the coefficients equal from both sides, we have:
- For \(x^3\): \(C = 0\),
- For \(x^2\): \(A + B = 0\),
- For \(x^1\): \(2A + 2C = 1\),
- For the constant term: \(2A = -6\).
From the last equation \(2A = -6\):
\[ A = -3. \]
Using \(A + B = 0\):
\[ -3 + B = 0 \Rightarrow B = 3. \]
Since \(C = 0\), we have determined:
\[ A = -3, \quad B = 3, \quad C = 0. \]
Thus, the respective values are:
\[ \boxed{-3}, \quad \boxed{3}, \quad \boxed{0}. \]
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