Question regarding a PDE I can see where to start on a $f(x,t)$ that solves a PDE $f_t+f*f_x+f_{xxx}=0$, then to show that it will also be true for $f(x-ct,t)+c$ for any number $c$
I can't see where to start on a problem. If a function $f(x,t)$ solves the PDE 
$$f_t + f f_x + f_{xxx} = 0$$
I want to show that $f(x-ct,t)+c$, for any number $c$, also solves the PDE.
 A: Hint: plug in the given expression in the LHS of the PDE and use the chain rule (and of course that $f$ solves the PDE)
EDIT: I can see that despite your efforts you haven't got to a solution, let me give you more insights:
Note that for $f(x-ct,t)+c$ one has, by the chain rule
$$\partial_t (f(x-ct,t))=-c\partial_xf(x-ct,t)+\partial_tf(x-ct,t)$$
Whereas the derivatives with respect to $x$ remain unchanged, i.e.
$$\partial_x(f(x-ct,t))=\partial_x f(x-ct,t).$$
Now plug these expressions in the left hand side:
\begin{align*}\partial_t & (f(x-ct,t))+(f(x-ct,t)+c)\partial_x(f(x-ct,t))+\partial_{xxx}(f(x-ct,t))\\ & =\color{red}{-c\partial_xf(x-ct,t)}+\partial_tf(x-ct,t)+(f(x-ct,t)+\color{red}{c)\partial_xf(x-ct,t)}+\partial_{xxx}f(x-ct,t)\\ & = \color{blue}{\partial_tf(x-ct,t)+f(x-ct,t)\partial_xf(x-ct,t)+\partial_{xxx}f(x-ct,t)}\\ & =0
\end{align*}
Where the two red terms cancel out because they are the same but of opposite sign, and the blue expression equals zero as $f$ solves the PDE at any point $(x_0,t_0)$, so in particular at $(x_0,t_0):=(x-ct,t)$.
A: I have started with putting the whole expression in like this 
$\frac{\partial (f(x-ct,t)+c)}{\partial t}+f(x,t) \frac{\partial (f(x-ct,t)+c)}{\partial x}+\frac{\partial (f(x-ct,t)+c)^3}{\partial^3 x}$  and then get $(-c,0)+(f(x-ct)+c)+(1,0) +0$ but it doesn't seme right?
A: With the chain rule I get $X(x)T'(t)+X(x)T(t)X'(x)T(t)+X'''(x)T(t)=0$
with $X(x)=x-ct$ and $T(t)=t$, where would $+c$ go? 
