Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$ 
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.

I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?

Multiply your expression by
$$\frac{  \sqrt{a^2+a+1}+\sqrt{a^2-a+1} } {\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}$$
To get
$$\frac{2a}{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1}}.$$
For $a$ positive, divide top and bottom by $a$ and push
the $1/a$ inside the square roots:
$$\frac{2}{ \sqrt{ 1+ \frac{1}{a} +\frac{1}{a^2} } + 
\sqrt{ 1 - \frac{1}{a} +\frac{1}{a^2} }}.$$
Now you can see that as $a$ gets very large, the above
fraction gets close to
$$\frac{2}{1+1} = 1.$$
For $a$ negative, do the same thing, but when you push the $1/a$ inside the square roots, you have to leave the negative outside.  The expression tends to $-1$.
Also, looking at the last expression, you can work out that the denominator is larger than the absolute value of the numerator, so the range is $(-1,1)$.
