$
\begin{align}
(x+3)(x+5)(x+7)(x+9)=9
\end{align}
$
Let $u = x+6$.
$
\begin{align}
(x+3)(x+5)(x+7)(x+9)=9
\\
(u-3)(u-1)(u+1)(u+3)=9
\\
(u-1)(u+1)(u-3)(u+3)=9
\\
(u^2-1)(u^2-9)=9
\end{align}
$
Let $z = u^2-5$.
$
\begin{align}
(u^2-1)(u^2-9)=9
\\
(z-4)(z+4)=9
\\
(z^2-16)=9
\end{align}
$
Now we can work our way backwards and solve for $x$.
$
\begin{align}
z^2=9+16
\\
z=\pm\sqrt{25}
\\
u^2-5=\pm\sqrt{25}
\\
u=\pm\sqrt{5\pm\sqrt{25}}
\\
x+6=\pm\sqrt{5\pm\sqrt{25}}
\\
x=\pm\sqrt{5\pm\sqrt{25}}-6
\end{align}
$
We find three solutions, $x=-6$ (double root), $x=\sqrt{10}-6$ and $x=-\sqrt{10}-6$.
1. $x=\sqrt{5-\sqrt{25}}-6=\sqrt{0}-6=-6$
2. $x=\sqrt{5+\sqrt{25}}-6=\sqrt{10}-6$
3. $x=-\sqrt{5-\sqrt{25}}-6=-\sqrt{0}-6=-6$
4. $x=-\sqrt{5+\sqrt{25}}-6=-\sqrt{10}-6$
<!-- basicblock-start oid="ObslEiK8sLJ7FHq1RHzmHqxu" -->
Solve $(x+3)(x+5)(x+7)(x+9)=9$.
::
Hint: Use symmetrical substitution.
<!-- basicblock-end -->