<s>[INST] Algebraic Proof help with real numbers Let $a$ and $b$ be distinct non-zero real numbers
Show that $-\frac{4ab}{(a-b)^2} = 1 - (\frac{a+b}{a-b})^2$
It's been 15 years since I have been in high school but I came across this problem and wanted to solve it.
$-\frac{4ab}{(a-b)^2} = -\frac{4ab}{a^2+b^2-2ab}$
And now I am stuck I assume I have to split this fraction up but I can't remember how to d that correctly. Any help will be appreciated.
 [/INST] Recall that
$(a+b)^2=a^2+2ab+b^2\tag{1}$$(a-b)^2=a^2-2ab+b^2\tag{2}$
Subtracting $(1)$ from $(2)$
$$4ab=(a+b)^2-(a-b)^2$$
$$\begin{align}-\dfrac{4ab}{(a-b)^2}&=-\dfrac{(a+b)^2-(a-b)^2}{(a-b)^2}\\&=-\left(\dfrac{a+b}{a-b}\right)+\dfrac{(a-b)^2}{(a-b)^2}\\&=1-\left(\dfrac{a+b}{a-b}\right)^2\qquad\blacksquare\quad\text{Q.E.D}\end{align}$$
</s>