Question

What are the respective values of A, B and C, so that x-6/X³+2x = a/x +bx+c/x² +2?

Answer 1

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:

1. Expand \( A(x^2 + 2) \):
\[
Ax^2 + 2A.
\]

2. 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:

1. For \(x^3\): \(C = 0\),
2. For \(x^2\): \(A + B = 0\),
3. For \(x^1\): \(2A + 2C = 1\),
4. 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:

\[
\boxed{-3}, \quad \boxed{3}, \quad \boxed{0}.
\]