Question

$find \: a \: natural \: number \: \: n \: \: such \: tht \: {3}^{9} + {3}^{12} + {3}^{15} + {3}^{n} \: is \: a \: perfect \: cube \: of \: an \: integer$
##No Useless answer plz.##

**Ans with steps**

Answer 1

Given Equation is 3^9 + 3^12 + 3^15 + 3^n.

It can be written as,

$(3)^3 + 3 * (3^3)^2 * 3^5 + (3^5)^3 + 3^n$

It is in the form of a^3 + 3a^2b + b^3 + 3ab^2.

Here,

a = 3^3

b = 3^5

3^n = 3ab^2

3^n = 3 * 3^3 * (3^5)^2

3^n = 3^14

n = 14.


Therefore the natural number = 14.


Note: I Took help from someone. Sorry about that.


Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks