Question

If $$\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$$
Show that: $$m-n=1$$

Answer 1

$$\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$$

$$\Rightarrow \dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$$

$$\Rightarrow \dfrac{3^{3n}.3^2-3^{3n}}{3^{3m}.2^3}=\dfrac{1}{3^3}$$

$$\Rightarrow \dfrac{3^{3n}(3^2-1)}{3^3m\times 8}=\dfrac{1}{3^3}$$

$$\Rightarrow \dfrac{3^{3n}\times 8}{3^3m\times 8}=\dfrac{1}{3^3}$$

$$\Rightarrow \dfrac{1}{3^{3(m-n)}}=\dfrac{1}{3^{3-1}}$$

$$\Rightarrow m-n=1$$ (proved)