Which solution to the equation $$ \frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^{2}-4} $$ is extraneous?

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We start with the equation

$$
\frac{3}{a+2}+\frac{2}{a}=\frac{4a-4}{a^{2}-4}.
$$

Notice that the denominator on the right factors as

$$
a^{2}-4=(a+2)(a-2).
$$

Therefore, the equation can be rewritten as

$$
\frac{3}{a+2}+\frac{2}{a}=\frac{4a-4}{(a+2)(a-2)}.
$$

Before solving, note the restrictions from the original denominators: 
- $$a+2\neq 0$$ implies $$a\neq -2$$,
- $$a\neq 0$$ from the second term,
- $$a^2-4\neq 0$$ implies $$a\neq 2$$ and $$a\neq -2$$.

Thus, $$a\neq -2$$, $$a\neq 0$$, and $$a\neq 2$$.

To eliminate the denominators, multiply both sides by the least common denominator $$\,a(a+2)(a-2)$$:

$$
a(a+2)(a-2)\left(\frac{3}{a+2}+\frac{2}{a}\right) = a(a+2)(a-2)\frac{4a-4}{(a+2)(a-2)}.
$$

Multiplying term-by-term:

- The first term:

$$
  \frac{3}{a+2}\cdot a(a+2)(a-2)=3a(a-2).
  $$
  
- The second term:

$$
  \frac{2}{a}\cdot a(a+2)(a-2)=2(a+2)(a-2).
  $$
  
- The right side simplifies to:

$$
  \frac{4a-4}{(a+2)(a-2)}\cdot a(a+2)(a-2)=a(4a-4).
  $$

Thus, the equation becomes

$$
3a(a-2)+2(a+2)(a-2)=a(4a-4).
$$

Next, expand each term:

1. Expand $$3a(a-2)$$:

$$
   3a(a-2) = 3a^2 - 6a.
   $$

2. Expand $$2(a+2)(a-2)$$. Recognize that $$(a+2)(a-2)=a^2-4$$:

$$
   2(a^2-4)=2a^2-8.
   $$

Adding these, the left-hand side becomes

$$
3a^2 - 6a + 2a^2 - 8 = 5a^2 - 6a - 8.
$$

The right-hand side is expanded as

$$
a(4a-4)=4a^2-4a.
$$

Now, set the equation:

$$
5a^2 - 6a - 8 = 4a^2 - 4a.
$$

Subtract $$4a^2-4a$$ from both sides:

$$
5a^2 - 6a - 8 - 4a^2 + 4a = 0.
$$

Combine like terms:

$$
a^2 - 2a - 8 = 0.
$$

This is a quadratic equation. Solve it using the quadratic formula $$a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$, with $$a=1$$, $$b=-2$$, and $$c=-8$$:

$$
a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4+32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}.
$$

Thus, the solutions are

$$
a = \frac{2+6}{2}=4 \quad \text{and} \quad a = \frac{2-6}{2}=-2.
$$

However, recall the restrictions on $$a$$. Since $$a$$ cannot be $$-2$$, the solution $$a=-2$$ is extraneous.

The extraneous solution is $$\boxed{-2}$$.
The extraneous solution is $$ a = -2 $$.

Which solution to the equation \( \frac{3}{a+2}+\frac{2}{a}=\frac{4 a-4}{a^{2}-4} \) is extraneous?

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