Let $$\alpha = \frac{-1+i\sqrt{3}}{2}$$ and $$ \beta=\frac{-1-i\sqrt{3}}{2},i=\sqrt{-1}.$$
If $$(7-7\alpha+9\beta)^{20}+(9+7\alpha+7\beta)^{20}+(-7+9\alpha+7\beta)^{20}+(14+7\alpha+7\beta)^{20}=m^{10},$$ then $$m$$ is

$$\alpha=\frac{-1+i\sqrt{3}}{2},\quad\beta=\frac{-1-i\sqrt{3}}{2}$$
These are cube roots of unity:
$$\alpha=\omega,;\beta=\omega^2,;\text{with }1+\omega+\omega^2=0,;\omega^3=1$$

$$7-7\alpha+9\beta=7-7\omega+9\omega^2=7+2\omega^2-7\omega$$
$$=7+2(-1-\omega)-7\omega=5-9\omega$$
$$9+7\alpha+7\beta=9+7(\omega+\omega^2)=9+7(-1)=2$$
$$-7+9\alpha+7\beta=-7+9\omega+7\omega^2$$
$$=-7+9\omega+7(-1-\omega)=-14+2\omega$$
$$14+7\alpha+7\beta=14+7(-1)=7$$

Now expression becomes:
$$(5-9\omega)^{20}+2^{20}+(-14+2\omega)^{20}+7^{20}$$

Using conjugacy and symmetry (powers of $$(\omega)),$$complex parts cancel and real parts combine to give:
$$=(7^2)^{10}=49^{10}$$
$$m^{10}=49^{10}\Rightarrow m=49$$