Question

. What is the natural number n for which

${3}^{9} + {3}^{12} + {3}^{15} + {3}^{n}$
is a perfect cube of an integer?

(A) 10 (B)11 (C)13 (D) 14​

Answer 1

Step-by-step explanation:

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.