Simplify and find the domain

Question

Simplify the following expression and find its domain

$$\frac{3a + 6b}{a-b}:\frac{a+2b}{2ab-2b^2}.$$

A. $6b\ \text{for}\ a \neq b\land a \neq -2b$

B. $6b\ \text{for}\ a \neq b\land (b \neq 0 \lor a \neq b)$

C. $6b\ \text{for}\ a \neq b\land (b \neq 0 \lor a \neq b) \land a \neq -2b$

D. $6b\ \text{for any}\ a, b$

E. $\dfrac{6b(1 - b)}{a-b}\ \text{for}\ a \neq b$

After simplification, I get $\mathbf{6b}$, but as for domain, none of the given answers seems correct.

Here are my calculations:

$$a - b \neq 0 \implies \mathbf{a \neq b}$$

$$\frac{a+2b}{2ab-2b^2} \neq 0 \implies a+2b \neq 0 \land 2ab - 2b^2 \neq 0$$

$$a + 2b \neq 0 \implies \mathbf{a \neq -2b}$$

$$2ab - 2b^2 = 2b(a-b)\neq 0 \implies \mathbf{b \neq 0 \land a \neq b}$$

So my answer is $\mathbf{6b\ \text{for}\ a \neq b \land a \neq -2b \land b \neq 0}$ and it seems similar to C, but after simplfication C becomes the same as A.

$$(a \neq b\land (b \neq 0 \lor a \neq b) \land a \neq -2b) = (a \neq b \land a \neq -2b),$$

because

$$(A \land (B \lor A) \land C) = (((A \land B) \lor (A \land A)) \land C) = (A \land C).$$

Do you have any idea where the problem is?

Answer 1

It's $$\frac{3(a+2b)}{a-b}:\frac{a+2b}{2b(a-b)}=6b,$$ which gives $$a\neq-2b\land a\neq b\land b\neq0,$$ which says that you are right and your book is wrong.

Answer 2

You are closed right. But note that the origin can write as $$\frac{3a+6b}{a-b}\frac{2ab-b^2}{a+2b},$$ so the domain is $a\neq b \land a\neq -2b$.