Some time ago, on this board I saw this approximation:
$\displaystyle \sqrt{x} \approx \frac{x+y}{2 \sqrt{y}}$
where $y$ is the nearest square number to $x$
It comes from expanding
$\left ( \sqrt{x}- \sqrt{y} \right )^2$
then setting
$\left ( \sqrt{x}- \sqrt{y} \right )=0$
and solving for $\sqrt{x}$
This gives the same result as the first two terms of the Taylor expansion
for $\sqrt{x}$ about $y$

It occurred to me that a similar process, expanding instead
$\left ( \sqrt{x}- \sqrt{y} \right )^3$
could lead to the first three terms of the Taylor expansion above.
I was wrong -- it gives a different, better result:
$\displaystyle \sqrt{x} \approx \frac{(3x+y) \sqrt{y}}{x+3y}$