Solve the equation : $2013x+\sqrt[4]{(1-x )^7}=\sqrt[4]{(1+x )^7}$. Solve the equation : $2013x+\sqrt[4]{(1-x )^7}=\sqrt[4]{(1+x )^7}$.
Show that it has percisely one root: $x=0$. 

$x=0$ is an obvious root and we need to restrict $x$ to  $|x|\le 1$ for the roots to make (real) sense in the first place.
First proceed with standard techniques, i.e. rearrange:
$$ 2013 x=\sqrt[4]{(1+x)^7}-\sqrt[4]{(1-x)^7}$$
Multiply by the conjugate:
$$ 2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)=\sqrt[2]{(1+x)^7}-\sqrt[2]{(1-x)^7}$$
Multiply by the conjugate again:
$$ 2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)\left(\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\right)={(1+x)^7}-{(1-x)^7}.$$
Since $|x|\le 1$, twe can estimate 
$$ \begin{align}\left|{(1+x)^7}-{(1-x)^7}\right|&=2\left|{7\choose 1}x+{7\choose 3}x^3+{7\choose 5}x^5+{7\choose 7}x^7\right|\\
&=2|x|\cdot \left|7+35x^2+21x^4+x^6\right|\\&\le128|x|. \end{align}$$
On the other hand, at least one of $1+x$, $1-x$ is $\ge 1$, hence  $\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\ge1$ and  $\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\ge1$ so that 
$$ \left|2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)\left(\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\right)\right|\ge 2013|x|.$$
In combniation, this gives us $2013|x|\le 128|x|$, i.e. $x=0$.
