Prove the divisbility of the following numbers.

(12 to the power of 8) x (9 to the power of 12) is divisible by 6 to the power of 16

blank x 6 to the power of 16

answer should be an exponent with a base

3^16

Step-by-step explanation:

12^8 * 9^12

Rewrite 12 as 3*4  and 9 as 3*3

(3*4)^8  * (3*3)^12

We know that (ab)^c = a^c * b^c

3^8  4^8  3^12  3^12

We can write 4 as 2^2

3^8  2^2^8  3^12  3^12

We know a^b^c = a^(b*c)

3^8  2^(2*8)  3^12  3^12

3^8  2^(16)  3^12  3^12

We also know that a^b *a^c *a^d = a^(b+c+d)

2^(16)   3^8   3^12  3^12

2^(16)  3^(8+12+12)

2^16  3^(32)

But we need 6 ^16 so we will need a 3^16  3^32 = 3^16 * ^16  (16+16=32)

2^16 *3^16 *3^16

Remember a^b*c^b = (ac)^b

(2*3)^16 3^16

6^16  3^16

3x^3-x^2+6x-9