Simplify ((a^2 - 2a + 4)/(a + 2))^2 - ((a^2 + 2a + 4)/(a - 2))^2.

Step-by-step solution

1. 1

   Spot the cube identities hiding in the numerators. Both numerators are the quadratic cofactors of a cube:

   $a^3+8=(a+2)(a^2-2a+4),\qquad a^3-8=(a-2)(a^2+2a+4)$

   So $\dfrac{a^2-2a+4}{a+2}=\dfrac{a^3+8}{(a+2)^2}$ and $\dfrac{a^2+2a+4}{a-2}=\dfrac{a^3-8}{(a-2)^2}$. Noticing this is what keeps the algebra from exploding.

2. 2

   Put both squares over the common denominator $(a^2-4)^2$. Since $(a+2)(a-2)=a^2-4$,

   $\left(\frac{a^2-2a+4}{a+2}\right)^{2}=\frac{(a^2-2a+4)^2(a-2)^2}{(a^2-4)^2},\qquad \left(\frac{a^2+2a+4}{a-2}\right)^{2}=\frac{(a^2+2a+4)^2(a+2)^2}{(a^2-4)^2}$

3. 3

   Subtract the numerators and expand. Both products are degree-6 polynomials; carrying out the multiplication and subtracting term by term gives

   $(a^2-2a+4)^2(a-2)^2-(a^2+2a+4)^2(a+2)^2=-16a^5-160a^3-256a$

   Every even power cancels — a useful sign that the two halves really are mirror images of each other under $a\mapsto -a$.

4. 4

   Factor the numerator completely.

   $-16a^5-160a^3-256a=-16a\bigl(a^4+10a^2+16\bigr)=-16a\,(a^2+2)(a^2+8)$

   Neither $a^2+2$ nor $a^2+8$ factors over the reals, so this is as far as it goes; note also that nothing cancels against $(a^2-4)^2$, so the fraction is already in lowest terms for $a\ne\pm 2$.

5. 5

   Check numerically at two values of $a$. At $a=1$ the original expression is $\left(\tfrac{3}{3}\right)^2-\left(\tfrac{7}{-1}\right)^2=1-49=-48$, and the closed form gives $\dfrac{-16(1)(3)(9)}{9}=-48$. At $a=3$ the original is $\left(\tfrac{7}{5}\right)^2-\left(\tfrac{19}{1}\right)^2=1.96-361=-359.04$, and the closed form gives $\dfrac{-16(3)(11)(17)}{25}=-\dfrac{8976}{25}=-359.04$. Both agree exactly.