Let the equation of a cubic polynomial

$P\left(x\right)=a{x}^3+b{x}^2+cx+d$

Now,

$P\left(1\right)=a+b+c+d=2$

$P\left(2\right)=8a+4b+2c+d=4$

$P\left(3\right)=27a+9b+3c+d=6$ $P\left(4\right)=64a+16b+4c+d=8\dots$

From Eqs. (i) and (ii), we get

$7a+3b+c=2$

From Eqs. (ii) and (iii), we get

$19 a+5 b+c=2$

From Eqs. (iii) and (iv), we get

$37 a+7 b+c=2$

Now, from Eqs. (v) and (vi), we get

$12a+2b=0$

and from Eqs. (vi) and (vii), we get

$18 a+2 b=0$

From Eqs. (viii) and (ix), we get

$\begin{array}{l}a=0\text{ and }b=0\\ c=2\text{ and }d=0\end{array}$

So, $P\left(x\right)=2 x$

$\therefore$ no cubic polynomial is possible.