Do you know the reverse process?

$\begin{aligned}f(\color{magenta}1)\Rightarrow&\ \color{#D61F06}a(\color{magenta}1)^3+\color{#20A900}b(\color{magenta}1)^2+\color{#3D99F6}c(\color{magenta}1)+\color{teal}d\Rightarrow &\color{#D61F06}a+\color{#20A900}b+\color{#3D99F6}c+\color{teal}d&=\ 10&\Rightarrow(1)\\f(\color{#302B94}2)\Rightarrow&\ \color{#D61F06}a(\color{#302B94}2)^3+\color{#20A900}b(\color{#302B94}2)^2+\color{#3D99F6}c(\color{#302B94}2)+\color{teal}d\Rightarrow&8\color{#D61F06}a+4\color{#20A900}b+2\color{#3D99F6}c+\color{teal}d&=\ 27&\Rightarrow(2)\\f(\color{#EC7300}3)\Rightarrow&\ \color{#D61F06}a(\color{#EC7300}3)^3+\color{#20A900}b(\color{#EC7300}3)^2+\color{#3D99F6}c(\color{#EC7300}3)+\color{teal}d\Rightarrow&27\color{#D61F06}a+9\color{#20A900}b+3\color{#3D99F6}c+\color{teal}d&=\ 60&\Rightarrow(3)\\f(\color{#624F41}4)\Rightarrow&\ \color{#D61F06}a(\color{#624F41}4)^3+\color{#20A900}b(\color{#624F41}4)^2+\color{#3D99F6}c(\color{#624F41}4)+\color{teal}d\Rightarrow&64\color{#D61F06}a+16\color{#20A900}b+4\color{#3D99F6}c+\color{teal}d&=\ 115&\Rightarrow(4)\end{aligned}$

$\begin{aligned}(2)-(1);\quad&7\color{#D61F06}a+3\color{#20A900}b+\color{#3D99F6}c&=\ 17&\Rightarrow(5)\\(3)-(2);\quad&19\color{#D61F06}a+5\color{#20A900}b+\color{#3D99F6}c&=\ 33&\Rightarrow(6)\\(4)-(3);\quad&37\color{#D61F06}a+7\color{#20A900}b+\color{#3D99F6}c&=\ 55&\Rightarrow(7)\end{aligned}$

$\begin{aligned}(6)-(5);\quad&12\color{#D61F06}a+2\color{#20A900}b&=\ 16&\Rightarrow(8)\\(7)-(6);\quad&18\color{#D61F06}a+2\color{#20A900}b&=\ 22&\Rightarrow(9)\end{aligned}$

$\begin{aligned}(9)-(8);\quad&6\color{#D61F06}a&=\ 6&\Rightarrow\color{#D61F06}a=\color{#D61F06}1\end{aligned}$

Instead $\color{#D61F06}a$ with $\color{#D61F06}1$ in equation $(8);$

$\begin{aligned}12(\color{#D61F06}1)+2\color{#20A900}b&=\ 16\\2\color{#20A900}b&=\ 4\Rightarrow\color{#20A900}b=\color{#20A900}2\end{aligned}$

Instead $\color{#D61F06}a$ with $\color{#D61F06}1$ and $\color{#20A900}b$ with $\color{#20A900}2$ in equation $(5);$

$\begin{aligned}7(\color{#D61F06}1)+3(\color{#20A900}2)+\color{#3D99F6}c&=\ 17\\13+\color{#3D99F6}c&=\ 17\Rightarrow\color{#3D99F6}c=\color{#3D99F6}4\end{aligned}$

Instead $\color{#D61F06}a$ with $\color{#D61F06}1$ , $\color{#20A900}b$ with $\color{#20A900}2$ and $\color{#3D99F6}c$ with $\color{#3D99F6}4$ in equation $(1);$

$\begin{aligned}\color{#D61F06}1+\color{#20A900}2+\color{#3D99F6}4+\color{teal}d&=\ 10\\7+\color{teal}d&=\ 10\Rightarrow \color{teal}d=\color{teal}3\end{aligned}$

Thus, $(\color{#D61F06}a\cdot\color{#20A900}b\cdot\color{#3D99F6}c\cdot\color{teal}d)+(\color{#D61F06}a+\color{#20A900}b+\color{#3D99F6}c+\color{teal}d)=(\color{#D61F06}1\cdot\color{#20A900}2\cdot\color{#3D99F6}4\cdot\color{teal}3)+10=24+10=\boxed{34}$

In matrix, the equation system is as follows:

$\begin{aligned} \begin{pmatrix} 1^3 & 1^2 & 1 & 1 \\ 2^3 & 2^2 & 2 & 1 \\ 3^3 & 3^2 & 3 & 1 \\ 4^3 & 4^2 & 4 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} & = \begin{pmatrix} 10 \\ 27 \\ 60 \\ 115 \end{pmatrix} \\ \Rightarrow \begin{pmatrix} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} & = \begin{pmatrix} 10 \\ 27 \\ 60 \\ 115 \end{pmatrix} \\ \Rightarrow \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} & = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 10 \\ 27 \\ 60 \\ 115 \end{pmatrix} \\ & = \frac{1}{6} \begin{pmatrix}-1 & 3 & -3 & 1 \\ 9 & -24 & 21 & -6 \\ -26 & 57 & -42 & 11 \\ 24 & -36 & 24 & -6 \end{pmatrix} \begin{pmatrix} 10 \\ 27 \\ 60 \\ 115 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \\ 3 \end{pmatrix} \\ & \\ \Rightarrow abcd + a+b+c+d & = 24 + 10 = \boxed{34} \end{aligned}$

I used the following Excel spreadsheet to do the calculations.

We will use Lagrange interpolation discussed here

Let $f(x)=p(x-1)(x-2)(x-3)+q(x-1)(x-2)(x-4)+r(x-1)(x-3)(x-4)+s(x-2)(x-3)(x-4)$

Put $f(1)=10$ We get $s=\frac{-5}{3}$

Similarly putting $f(2)=27,f(3)=60,f(4)=115$ yields respectively the values

$r=\frac{27}{2},q=-30,p=\frac{115}{6}$

Put in $f(x)$ to get

$f(x)=\frac{115}{6}(x^3-6x^2+11x-6)-30(x^3-7x^2+14x-8)+\frac{27}{2}(x^3-8x^2+19x-12)-\frac{5}{3}(x^3-9x^2+26x-24)$

$f(x)=x^3+2x^2+4x+3$

Clearly $a=1,b=2,c=4,d=3$

$a.b.c.d = 24$ and $a+b+c+d=10$

$Answer = 24+10 = \boxed{34}$