To find the values of \( A \), \( B \), and \( C \) such that
\[ \frac{x - 6}{x^3 + 2x} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^2 + 2}, \]
we first simplify the left-hand side.
We can factor the denominator of the left side:
\[ x^3 + 2x = x(x^2 + 2). \]
So we have:
\[ \frac{x - 6}{x(x^2 + 2)}. \]
Next, rewrite the right-hand side with a common denominator:
The common denominator for the right-hand side is \( x^2(x^2 + 2) \). Thus, we rewrite the fractions:
\[ \frac{A}{x} = \frac{A \cdot x(x^2 + 2)}{x^2(x^2 + 2)} = \frac{Ax(x^2 + 2)}{x^2(x^2 + 2)}, \] \[ \frac{B}{x^2} = \frac{B \cdot (x^2 + 2)}{x^2(x^2 + 2)} = \frac{B(x^2 + 2)}{x^2(x^2 + 2)}, \] \[ \frac{C}{x^2 + 2} = \frac{C \cdot x^2}{x^2(x^2 + 2)} = \frac{Cx^2}{x^2(x^2 + 2)}. \]
Now combine these fractions:
\[ \frac{Ax(x^2 + 2) + B(x^2 + 2) + Cx^2}{x^2(x^2 + 2)}. \]
Equating both sides, we have:
\[ \frac{x - 6}{x(x^2 + 2)} = \frac{Ax(x^2 + 2) + B(x^2 + 2) + Cx^2}{x^2(x^2 + 2)}. \]
Multiplying both sides by \( x^2(x^2 + 2) \) leads to:
\[ x - 6 = Ax(x^2 + 2) + B(x^2 + 2) + Cx^2. \]
We now expand the right side:
\[ Ax^3 + 2Ax + Bx^2 + 2B + Cx^2 = Ax^3 + (B + C)x^2 + 2Ax + 2B. \]
We can now set coefficients from both sides of the equation. For the left side \( x - 6 \), we have:
- Coefficient of \( x^3 \): \( A = 0 \)
- Coefficient of \( x^2 \): \( B + C = 0 \)
- Coefficient of \( x \): \( 2A = 1 \) (since \( A = 0 \) this doesn't apply)
- Constant term: \( 2B = -6 \)
From \( 2B = -6 \):
\[ B = -3. \]
Using \( B + C = 0 \):
\[ -3 + C = 0 \Rightarrow C = 3. \]
Now, substituting back into our equation, we have:
\[ A = 0, \quad B = -3, \quad C = 3. \]
Thus, the values of \( A \), \( B \), and \( C \) are:
\[ \boxed{(0, -3, 3)}. \]
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