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: $x=0$. 
 A: Let $f(x)=2013x+\sqrt[4]{(1-x )^7}-\sqrt[4]{(1+x )^7}$. We want to solve the equation $f(x)=0$. Observe that the domain of $f$ is $[-1,1]$. Furthermore, its derivative is:
$$
f'(x)=2013+ \dfrac{7 \sqrt[4]{-(x-1)^7}}{4 (x-1)}-\dfrac{7 (x+1)^6}{4 ((x+1)^7)^{3/4}}
$$
which is positive for all $x\in(-1,1)$. Hence, $f$ is monotone increasing, so $x=0$ is its only root.
A: $x=0$ 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. $x=0$.
A: Set : $u=\sqrt[4]{1-x}\Rightarrow u^{4}=1-x,v=\sqrt[4]{1+x}\Rightarrow v^{4}=1+x$.
We have $\begin{cases}u^{4}+v^{4}=2\\2013\left(v^{4}-u^{4} \right)+2u^{7}=2v^{7} \tag{2} \end{cases}$
from $(2)$ : $\Rightarrow 2013v^{4}-2v^{7}=2013u^{4}-2u^{7}$
with $f(t)=2013t^4-2t^7$ ,
$f'(t)=t^{3}\left(8052-14t^{3} \right)\geq 0$
$\Rightarrow u=v\Leftrightarrow u^{4}=1\iff u=1\iff x=0$
