If $(1+\mathrm{x}{)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_1\mathrm{x}{+}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{x}}^2+\dots .{+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ where ${ }^{\mathrm{n}}{\mathrm{C}}_0,{ }^{\mathrm{n}}{\mathrm{C}}_1,{ }^{\mathrm{n}}{\mathrm{C}}_2,\dots$ are binomial coefficients, then $2\left({\mathrm{C}}_0+{\mathrm{C}}_3+{\mathrm{C}}_6+\dots .\right)+\left({\mathrm{C}}_1+{\mathrm{C}}_4+{\mathrm{C}}_7+\dots \dots \right)(1+\mathrm{\omega })+\left({\mathrm{C}}_2+{\mathrm{C}}_5+{\mathrm{C}}_8+\dots ..\right)\left(1+{\mathrm{\omega }}^2\right)$, where $\mathrm{\omega }$ is the cube root of unity and n is a multiple of 3, then the above expression is equal to