ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. 
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$

My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
 A: Now you use the rational root theorem, Horner's algorithm and a little work to find the roots of the polynomial.
The rational candidates for the root are $\frac{p}{q}$ where $p$ is a factor of $225$ (i.e., $3^2\cdot 5^2$) and $q$ is a factor of $104$ (i.e., $2^3\cdot 13$).
A: Let $u=x$ and $v=3-x$. Then we have
$$
u+v = 3,
\qquad
\frac{1}{u^2} + \frac{1}{v^2} = \frac{104}{25}
$$
But
$$
\frac{1}{u^2} + \frac{1}{v^2}
=
\frac{u^2+v^2}{(uv)^2}
=
\frac{(u+v)^2-2uv}{(uv)^2}
=
\frac{9-2uv}{(uv)^2}
$$
Thus, $uv$ is a root of
$$
\frac{9-2z}{z^2} = \frac{104}{25}
$$
a quadratic equation. Once you know $uv$ and $u+v$, you know $u$ and $v$ by solving another quadratic equation.
A: Let $y:=x-\dfrac{3}{2}$.  The equation becomes
$$\frac{1}{\left(y+\frac{3}{2}\right)^2}+\frac{1}{\left(y-\frac{3}{2}\right)^2}=\frac{104}{25}\,.$$
This is equivalent to
$$\frac{y^2+\frac{9}{4}}{\left(y^2-\frac{9}{4}\right)^2}=\frac{52}{25}\,.$$
Let $z:=\dfrac{1}{y^2-\frac{9}{4}}$, we have
$$\frac{9}{2}z^2+z=z^2\left(\frac{1}{z}+\frac{9}{2}\right)=\frac{52}{25}\,.$$
That is,
$$\frac{9}{2}\left(z+\frac{4}{5}\right)\left(z-\frac{26}{45}\right)=0\,.$$
Thus, $z=-\dfrac45$ or $z=\dfrac{26}{45}$.
In the case $z=-\dfrac{4}{5}$, we have
$$y^2-\frac{9}{4}=\frac{1}{z}=-\frac{5}{4}\,,$$
so $y^2=1$, or $y=\pm1$.  In this case, $x=\dfrac{1}{2}$ or $x=\dfrac{5}{2}$.
In the case $z=\dfrac{26}{45}$, we have
$$y^2-\frac{9}{4}=\frac{1}{z}=\frac{45}{26}\,.$$
That is, $y^2=\dfrac{207}{52}$, so $y=\pm\dfrac{3\sqrt{299}}{26}$.  Hence,
$$x=\frac{39\pm3\sqrt{299}}{26}\,.$$
A: Now, you can make the following.
Easy to see that $\frac{1}{2}$ and $\frac{5}{2}$ they are roots of the equation,
which gives a factor $$(2x-1)(2x-5)=4x^2-12x+5$$ and
$$104x^4-624x^3+886x^2+150x-225=$$
$$=104x^2-312x^3+130x^2-312x^3+936x^2-390x-180x^2+540x-225=$$
$$=26x^2(4x^2-12x+5)-78(4x^2-12x+5)-45(4x^2-12x+5)=$$
$$=(4x^2-12x+5)(26x^2-78x-45),$$ which gives the answer:
$$\left\{\frac{1}{2},\frac{5}{2},\frac{3}{2}\left(1+\sqrt{\frac{23}{13}}\right),\frac{3}{2}\left(1-\sqrt{\frac{23}{13}}\right)\right\}$$
A: Due to the symmetry at $x=3/2$, it makes sense to let $x=(3-u)/2$, which simplifies things initially to
$${1\over(3-u)^2}+{1\over(3+u)^2}={26\over25}$$
and then to
$${9+u^2\over(9-u^2)^2}={13\over25}$$
This can now be turned into a quadratic in $U=u^2$: $25(9+U)=13(9-U)^2$, or, once you do the algebra, $13U^2-259U+828=0$, which factors into
$$(U-4)(13U-207)=0$$
On noting that $207=9\cdot29$, we see from this that $u=\sqrt U=\pm2$ and $\pm3\sqrt{29/13}$, so that from $x=(3-u)/2$ we get
$$x=1/2,\quad 5/2,\quad3(1-\sqrt{29/13})/2,\quad\text{and}\quad3(1+\sqrt{29/13})/2$$
Remarks: Knowing that the symmetry between $x$ and $3-x$ makes a substitution worth trying is mostly a matter of experience. Figuring out exactly what substitution makes life easiest is a combination of experience and fiddling around. The factorization of the quadratic is easiest if you notice that $25(9+4)=13(9-4)^2$; if you don't, then the quadratic formula (and a calculator to compute $\sqrt{259^2-4\cdot13\cdot828}$) will do the job.
A: One can see that $1/2$ is a solution of $\frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. Then it is easy to see that also $5/2$ is a solution.
Can you proceed ?
