# Simplify. `(3^(3n)-1)/(3^(n+1)-3)` (3to the power of (3n) -1 all on 3 to the power of (n+1) -3)

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### 1 Answer

`(3^(3n)-1)/(3^(n+1)-3)`

To simplify, at the numerator apply the exponent property `a^(xy)=(a^x)^y` .

`=((3^n)^3-1)/(3^(n+1)-3)`

Then, factor the top using using the factor of difference of two cubes which is `a^3-b^3=(a-b)(a^2+ab+b^2)` .

`=((3^n-1)((3^n)^2+3^n*1 + 1^2))/(3^(n+1)-3)`

`=((3^n-1)(3^(2n)+3^n+1^2))/(3^(n+1)-3)`

Since 3^2=9 and 1^2 =1 then:

`=((3^n-1)(9^n+3^n+1))/(3^(n+1)-3)`

Next, at the denominator apply the exponent property.`a^(x+y)=a^x*a^y` .

`=((3^n-1)(9^n+3^n+1))/(3^n*3^1-3)`

`=((3^n-1)(9^n+3^n+1))/(3*3^n-3)`

And, factor out the GCF.

`=((3^n-1)(9^n+3^n+1))/(3(3^n-1))`

Now that both the top and bottom are in factored form, cancel the common factor which is `3^n-1.`

`=(9^n + 3^n+1)/3`

**Hence, the simplified form of the given expression is `(9^n+3^n+1)/3` .**