Determine all $f:\Bbb R\to \Bbb R$ such that $f\big(a-3f(b)\big)=f\left(a+f(b)+b^3\right)+f\left(4f(b)+b^3\right)+1$ for every $a,b\in\Bbb R$. Im struggling with this functional equation:

Determine all $f: \Bbb R \to \Bbb R$ such that
$$f\big(a-3f(b)\big)=f\left(a+f(b)+b^3\right)+f\left(4f(b)+b^3\right)+1$$
for all $a,b\in\Bbb R$.

Clearly the constant function $f(x)=-1$ for all $x\in\mathbb{R}$ is a solution.  If $k=f(0)$, then $k\neq 0$.  Otherwise, if $a,b=0$, we have
$$f\big(-3f(0)\big)=f\big(f(0)\big)+f\big(4f(0)\big)+1\,,$$
which would give $0=0+0+1$ if $k=0$.  By taking $b=0$, we have
$$f(a-3k)=f(a+k)+f(4k)+1,$$
or
$$f(a+4k)=f(a)+r,$$
where $r=-f(4k)-1$.  This proves that
$$f(a+4kn)=f(a)+nr$$
for all integers $n$.  What to do next?
Thanks in advance for your time.
 A: Let $t\in\mathbb{R}$.  Suppose that $f:\Bbb R\to\Bbb R$ satisfies the functional equation
$$f\big(a-3f(b)\big)=f\big(a+f(b)+b^3\big)+f\big(4f(b)+b^3\big)+t\tag{1}$$
for all $a,b\in\Bbb{R}$.  Substituting $a+3f(b)$ for $a$ yields
$$f(a)=f\big(a+g(b)\big)+f\big(g(b)\big)+t,\tag{2}$$
where $g(b)=4f(b)+b^3$.  From $(2)$, when $a=0$, we have
$$f\big(g(b)\big)=\frac{f(0)-t}{2}.\tag{3}$$  This shows that
$$f\big(a+g(b)\big)=f\big(a+g(c)\big),$$
or equivalently
$$f\Big(a+\big(g(b)-g(c)\big)\Big)=f(a)$$
for all $a,b,c\in\Bbb R$.
Let $S=\big\{g(b):b\in\Bbb{R}\big\}$, $T=S-S=\big\{x-y:x,y\in S\big\}$, and $U=\operatorname{span}_{\Bbb Z}T$.  It follows that $$f(a+u)=f(a)\tag{4}$$ for every $a\in \Bbb R$ and $u\in U$.  Define $V=\operatorname{span}_{\Bbb Q}T$.  Then $V$ has a Hamel basis $\mathcal{H}$, and every element of $U\subseteq V$ is a linear combination of elements of $\mathcal{H}$.  In other words,
$$g(b)-g(0)=4\sum_{x\in\mathcal{H}}h_x(b)x\in U$$
for some functions $h_x:\Bbb R\to\Bbb Q$ such that for each $b\in\Bbb{R}$, $h_x(b)= 0$ for all but finitely many $x\in\mathcal{H}$.  Therefore,
$$g(b)=4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x$$
and
$$f(b)=-\frac{b^3}{4}+f(0)+\sum_{x\in\mathcal{H}}h_x(b)x.\tag{5}$$
Note that
$$\sum_{x\in\mathcal{H}}h_x(b)x\in\frac{1}{4}U=\left\{\frac{u}{4}:u\in U\right\}.$$
Using $(2)$ and $(3)$, we have
$$-\frac{a^3}{4}+f(0)+\sum_{x\in\mathcal{H}}h_x(a)x=-\frac{\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)^3}{4}+f(0)+\sum_{y\in\mathcal{H}}h_y\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)y+\frac{f(0)+t}{2}.$$
Therefore
$$\frac{\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)^3-a^3}{4}-\frac{f(0)+t}{2}=\sum_{y\in\mathcal{H}}h_y\left(a+4f(0)+4\sum_{x\in\mathcal{H}}h_x(b)x\right)y\in \frac{1}{4}U.$$
The left hand side is a continuous function in $a$ whose range is a connected subset of $\mathbb{R}$ (i.e., an interval).  Therefore, unless
$$f(0)+\sum_{x\in\mathcal{H}}h_x(b)x=0$$
for every $b\in\mathbb{R}$, we must have $U=\mathbb{R}$.  
If $f(0)+\sum_{x\in\mathcal{H}}h_x(b)x=0$ for all $b\in\mathbb{R}$, then by $(5) $ we have
$$f(b)=-\frac{b^3}{4}$$
for all $b\in\Bbb R$.  This can happen if and only if $t=0$.  We now assume that $U=\mathbb{R}$.
If $U=\mathbb{R}$, then by $(4)$, $f$ is constant.  Thus, from $(2)$, $$f(a)=-t$$ for every $a\in\mathbb{R}$.
In general, let $p\neq 0$ and $\gamma\neq -1$ be real numbers, and $q:\Bbb R\to\Bbb R$ a continuous function such that for every real number $\tau\neq 0$, the function $\Delta_\tau q:\mathbb{R}\to\mathbb{R}$ defined by $$\Delta_\tau q(z)=q(z+\tau)-q(z)$$ is not a constant function.  For example, $q$ may be a polynomial of degree at least $2$.  Then all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$f(a)=f\big(a+pf(b)+q(b)\big)+\gamma f\big(pf(b)+q(b)\big)+t$$
are


*

*the constant function $f(a)=-\frac{t}{\gamma}$ if $\gamma\neq 0$;

*the function $f(a)=-\frac{1}{p}q(a)$ if $t=\frac{\gamma}{p}q(0)$.


In the case $\gamma=0$ and $t\neq 0$, there are no solutions.
