Simplify the expression: $\frac{ 3^{2n} -1} { 3^{n+1}-3} $

Question

Factorizing indices is quite a cool and intriguing area of indices. However, I stumbled upon one that I couldn't do any further:

$$\frac{ 3^{2n} -1} { 3^{n+1}-3} $$

I got up to this stage:

$$\frac{ 3^{n} \cdot 3^{n} -3^0} { 3^{n} \cdot 3^1 -3^1} $$

I got to this stage through using the laws of indices and breaking them down. I feel that I'm close, but not there yet. Please help. Thank you!

Answer 1

$$\frac{ 3^{2n} -1} { 3^{n+1}-3}=\frac { { \left( { 3 }^{ n } \right) }^{ 2 }-{ 1 }^{ 2 } }{ 3\left( { 3 }^{ n }-1 \right) } =\frac { \left( { 3 }^{ n }-1 \right) \left( { 3 }^{ n }+1 \right) }{ 3\left( { 3 }^{ n }-1 \right) } =\frac { { 3 }^{ n }+1 }{ 3 } $$

Answer 2

The denominator can be broken down into $3(3^n-1)$. The numerator can also be broken down into $(3^n-1)(3^n+1)$. Simplifying this gives us $(3^n+1)/3$ as the answer.

Answer 3

$$\frac{3^{2n}-1}{3^{n+1}-3}=\frac{(3^n)^2-1^2}{3^n\cdot3-3}=\frac{(3^n+1)(3^n-1)}{3(3^n-1)}=\frac{3^n+1}3$$