The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$ How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$

All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$.  The functional equation becomes 
$g(x+y) = g(x) + g(y)$.  This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$.  Of course $c$ can depend on $z$.  Now we need to solve 
$f(x+z) - f(x) - f(z) = c(z) x$.   Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$. 
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$. 
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$.  Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even, 
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$.  now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$. 
