Simplify ((a − 2)/(a + 2) − (a + 2)/(a − 2)) ÷ (12a²/(4 − a²)).

Step-by-step solution

1. 1

   Combine the bracket over the common denominator (a+2)(a−2).

   $\frac{(a-2)^{2}-(a+2)^{2}}{(a+2)(a-2)}$

   The numerator is itself a difference of squares: $(A^{2}-B^{2})=(A-B)(A+B)$ with $A=a-2$, $B=a+2$ gives $(-4)(2a)=-8a$. (Expanding also works: $a^{2}-4a+4-a^{2}-4a-4=-8a$.)

2. 2

   Write the bracket compactly. Since $(a+2)(a-2)=a^{2}-4$,

   $\text{bracket}=\frac{-8a}{a^{2}-4}$

3. 3

   Turn the division into multiplication by the reciprocal.

   $\frac{-8a}{a^{2}-4}\cdot\frac{4-a^{2}}{12a^{2}}$

4. 4

   Use 4 − a² = −(a² − 4) to cancel the quadratics. The two sign-opposite factors cancel to $-1$:

   $\frac{-8a}{a^{2}-4}\cdot\frac{-(a^{2}-4)}{12a^{2}}=\frac{8a}{12a^{2}}$

   Missing this sign flip is the one place the answer can come out negative by mistake.

5. 5

   Reduce the remaining fraction. Cancel $4a$ from top and bottom:

   $\frac{8a}{12a^{2}}=\frac{2}{3a}$

   $\boxed{\dfrac{2}{3a}}\qquad(a\neq 0,\ a\neq\pm 2)$

6. 6

   Check with a value. At $a=1$: the bracket is $\tfrac{-1}{3}-\tfrac{3}{-1}=-\tfrac13+3=\tfrac83$, and the divisor is $\tfrac{12}{3}=4$, so the quotient is $\tfrac{8}{12}=\tfrac23$. The answer gives $\tfrac{2}{3(1)}=\tfrac23$ ✓.