Solution:

If we define the signals,

$ x_1(n)=2^n u(n)\\ $

$ x_2(n)=3^n u(n)\\ $

then x(n) can be written as

$ x(n)=3 x_1(n)-4 x_2(n)\\ $

According to (3.2.1), its z-transform is

$ X(z)=3 X_1(z)-4 X_2(z)\\ $

From (3.1.7) we recall that,

$ \alpha^n u(n) \stackrel{\vdots}{\longleftrightarrow} \frac{1}{1-\alpha z^{-1}} \quad \text { ROC: }|z|\gt|\alpha|\\ $

By setting $\alpha=2$ and $\alpha=3$ in (3.2.2). we obtain,

$ \begin{array}{ll}\\ x_1(n)=2^n u(n) \stackrel{\ddots}{\longleftrightarrow} X_1(z)=\frac{1}{1-2 z^{-1}} & \text { ROC }:|z|\gt2\\ \\ x_2(n)=3^n u(n) \stackrel{i}{\longleftrightarrow} X_2(z)=\frac{1}{1-3 z^{-1}} & \text { ROC: }|z|\gt3\\ \end{array}\\ $

The intersection of the ROC of $X_1(z)$ and $X_2(z)$ is $|z|\gt3$. Thus the overall transform $X(z)$ is,

$ X(\xi)=\frac{3}{1-2 z^{-1}}-\frac{4}{1-3 z^{-1}} \quad \mathrm{ROC}:|z|\gt3\\ $