This shouldn't be too hard, if you know the shortcut

Here let $a, b$ be alpha and beta.*

Let $g (x)=f (x)-2x$ for all real $x$ .

Now $g(x)=0$ for $x$ equal to $a, b$ and $a+b$ which is 2.

Thus $g (x)$ has factors $(x-a), (x-b), (x-2)$

Since none of $a, b, 2$ are equal, we get:

$g (x)=c (x-a)(x-b)(x-2)=c (x^{2}-2x-1)(x-2)$

Now when $x=ab=-1,$ substitute it to get $c=-1$ . Adding $2x$ results in

$f (x)=-(x^{2}-2x-1)(x-2)+2x$ . Substitute -2 and get the answer of 24.

*Sorry, but I don't know how to latex alpha and beta. I will do some research.

Here is another way. This is the shortcut that is mentioned in the title.

First rewrite the given statements like this: $f(\alpha )-2\alpha =0,\quad f(\beta )-2\beta =0,\quad f(\alpha +\beta )-2(\alpha +\beta )=0,\quad f(-1)=4$ $Define\quad g(x)=f(x)-2x$

Then, notice that $\alpha$ and $\beta$ and $\alpha+\beta$ are the 3 solutions of g(x). Since the problem states that f(x) is cubic, so is g(x). This allows us to assume that those 3 are the only solutions of g(x)=0. We can write g(x) like this: $g(x)=a(x-\alpha )(x-\beta )(x-(\alpha +\beta ))=a(x-\alpha )(x-\beta )(x-2)=f(x)-2x\quad (a\neq 0)\\ \therefore f(x)=2x+a(x-\alpha )(x-\beta )(x-2)=2x+a[x^{ 3 }-(\alpha +\beta +2)x^{ 2 }+\{ \alpha \beta +2(\alpha +\beta )\} x-2\alpha \beta ]\\ =2x+a\{ x^{ 3 }-4x^{ 2 }+3x+2\} =ax^{ 3 }-4ax^{ 2 }+(3a+2)x+2a$ Now, we just have to find 'a'. We do that by using the fact that $f(-1)=4$ $f(-1)=a(-1)^{ 3 }-4a(-1)^{ 2 }+(3a+2)(-1)+2a=-a-4a-3a-2+2a=-6a-2=4\\ \therefore a=-1\rightarrow f(x)=-x^3+4x^2-x-2$

Therefore, $f(-2)=-(-2)^{ 3 }+4(-2)^{ 2 }-(-2)-2=24$

The roots of $\space x^2-2x-1=0\space$ are:

$(\alpha, \beta ) = \frac {2\pm\sqrt{4+4}}{2} = 1 \pm \sqrt{2}\quad \Rightarrow \alpha + \beta = 2 \quad \Rightarrow \alpha \beta = -1$

Let $\space f(x) = ax^3 + bx^2 + cx + d \space$ , then we have:

$\begin{cases} f(\alpha) & = & a(\alpha)^3+b(\alpha)^2+c(\alpha)+d & = & 2\alpha \\ f(\beta) & = & a(\beta)^3+b(\beta)^2+c(\beta)+d & = & 2\beta \\ f(\alpha+\beta) & = & a(\alpha+\beta)^3+b(\alpha+\beta)^2+c(\alpha+\beta)+d & = & 2\alpha + 2\beta \\ f(\alpha\beta) & = & a(\alpha\beta)^3+b(\alpha\beta)^2+c(\alpha\beta)+d & = & 4 \end{cases}$

$\begin{cases} f(1+\sqrt{2}) & = & (7+5\sqrt{2})a+(3+2\sqrt{2})b+(1+\sqrt{2})c+d & = & 2+2\sqrt{2} \\ f(1-\sqrt{2}) & = & (7-5\sqrt{2})a+(3-2\sqrt{2})b+(1-\sqrt{2})c+d & = & 2-2\sqrt{2} \\ f(2) & = & 8a+4b+2c+d & = & 4\\ f(-1) & = & -a+b-c+d & = & 4 \end{cases}$

$\begin{cases} Eq.1: f(\alpha) - f(\beta) & = & 10\sqrt{2}a+4\sqrt{2}b+2\sqrt{2}c & = & 4\sqrt{2} \\ Eq.2: f(2)-f(-1) & = & 9a+3b+3c & = & 0\\ Eq.3: f(\alpha) - f(-1) & = & (8+5\sqrt{2})a+(2+2\sqrt{2})b+(2+\sqrt{2})c & = & -2+2\sqrt{2} \end{cases}$

$\begin{cases} Eq.4: \frac {Eq.1}{2\sqrt{2}} & \Rightarrow & 5a+2b+c & = & 2 \\ Eq.5: \frac {Eq.2}{3} & \Rightarrow & 3a+b+c & = & 0\\ Eq.6: \frac {1}{2}(Eq.3 - \frac {1}{2}Eq.1) & \Rightarrow & 4a+b+c & = & -1 \end{cases}$

$\begin{cases} Eq.7: Eq.6-Eq.5 & \Rightarrow & a & = & -1 \\ Eq.8: Eq.4-Eq.5 & \Rightarrow & 2a+b & = & 2 \quad \Rightarrow b = 4 \\ Eq.5: & \Rightarrow & 3a+b+c & = & 0 \quad \Rightarrow c = -1 \\ f(-1) & = & -a+b-c+d & = & 4 \quad \Rightarrow d = -2 \end{cases}$

Therefore, $f(-2) = (-1)(-8)+4(4)+(-1)(-2)-2 = 8+16+2-2 = \boxed{24}$

From the first three conditions, we observe that the remainder is twice the value taken by the function. As it is a cubic polynomial, we can construct it as:

k[(x-a)(x-b){x-(a+b)}] + 2x

Now a=1+sqrt(2), b=1-sqrt(2).

We know ab=-1

By putting this in the above constructed cubic polynomial, we get value of k=-1.

Now we simply plug in -2 and get the answer as ${24}$