3293136(x) = an integer to the third power. What is the least positive integer that x could be to satisfy this

Answer 1

3293136 = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 3 * 3 * 7 * 11 * 11

So you would need each factor to appear 3 times (or multiples of 3) to make it a power of 3.

(2 * 2 * 2) * (2 * 2 * 2) * (3 * 3 * 3) * (3 * 3 * 3) * (7 * 7 * 7) * (11 * 11 * 11)

So altogether you add the factors 2 * 2 * 3 * 7 * 7 * 11 = 6468

x = 6468

To double-check:
3,293,136 * 6,468 = 21,300,003,648 = 2,772^3

Answer 2

Well, first factor 3293136.... (205821 * 16) = ( 9 * 22869 * 4 * 4) = ( 9 * 9 * 4 * 4 * 2541) = (9 * 9 * 4 * 4 * 3 * 847) = (9 * 9 * 4 * 4 * 3 * 7 * 121) = ( 9 * 9 * 4 * 4 * 3 * 7 * 11 * 11)

Well, you can do the factorization any way you want, thats how I did it, I kept the 9's and 4's even though they can be factored down because its easier in my mind.

So, if you have ( 9 * 9 * 4 * 4 * 3 * 7 *11 * 11) how can you make it something thats to the 3rd power? Each number in there needs to be able to be grouped up in 3's. So, if you multiply by 3, the 3 there can group with it to make the 3rd 9. Then you need to multiply by 4 as well, to get 4 of them, and 11 to get 3 of them, then you need 2 7's to make them a pair.

So, in order to get the least positive integer to make it an integer to the 3rd power, x must equal 3 * 4 * 7 * 7 *11 =6468

3293136 *6468 = 21300003648 = 2772 ^ 3

Answer 3

first, factor out the largest integer with perfect cube, in this case it's 216. divide the coefficient by 216 you get:

15246

get the prime factors:

2,3^2,7,11^2

with the prime factors as base, get the exponent value that will result to a perfect cube, you get:
2^2,3,7^2,11

multiply them altogether: 2^2 x 3 x 7^2 x 11 = 6468

ans: 6468

hope this helps