Olympiad-like way to solve the equation $(x^2-4)(x^2+6x+6)=x^2-1$? 
Solve the equation:
$$(x^2-4)(x^2+6x+6)=x^2-1$$

I found this question from math olympiad textbook for beginners. But there is no specific hint for the solution.
Is there any faster way to solve this equation?
I see that the given equation is equivalent to
$$(x-2)(x+2)(x^2+6x+6)=(x-1)(x+1)$$
But, I don't see how can I proceed.
After expanding I got
$$x^4+6x^3+x^2-24x-23=0$$
Now, I need factorisation. But factoring doesn't seem like the good track to me.
I know that the substitution $x=y-\frac {b}{4a}$ in $ax^4+bx^3+cx^2+dx+e=0$ can work. But, this gets us a lot of more work. Because the original equation was not given that way.
 A: We can use the Tschirnhaus trasnsformation:

*

*Consider the change of variables $u: x\mapsto \frac{\color{blue}{6}}{4}+x$, then the equation $x^{4}+\color{blue}{6}x^{3}+x^{2}-24x-23=0$ can be written as $$u^{4}-\frac{25}{2}u^{2}+\frac{1}{16}=0.$$

*Another, change of variables $v:u\mapsto u^{2} $ give the equation $$v^{2}-\frac{25}{2}v+\frac{1}{16}=0$$

*Now, just use the fact $$v=\frac{\frac{25}{2}\pm \sqrt{\left(\frac{25}{2}\right)^{2}-4\left(\frac{1}{16}\right)}}{2}.$$

*Finally, substitution back.

A: We want to make the right-hand side a constant. Therefore, we subtract $x^2-4$ from both sides:
$$
\begin{align}(x^2-4)(x^2+6x+6)-(x^2-4)=(x^2-1)-(x^2-4)\end{align}
$$
Thus we obtain:
$$(x^2-4)(x^2+6x+5)=3$$
The following steps will lead us to the solution:
$$
\begin{align}
&(x-2)(x+2)(x+1)(x+5)=3\\
\implies &\color{#E2062C}{(x-2)(x+5)}\color{#0000CD}{(x+1)(x+2)}=3\\
\implies &(\color{#E2062C}{x^2+3x}-10)(\color{#0000CD}{x^2+3x}+2)=3\end{align}
$$
Then letting $x^2+3x=u$, you have:
$$
\begin{align}&(u-10)(u+2)=3\\
\implies &u^2-8u-23=0.\end{align}
$$
Finally, you can use the quadratic formula to complete the solution.
