Given that the product of the x-th power of 2 by the y-th power of 3 by the z-th power of 111 is equal to 1998, where x.y.z is a natural number, how to find the value of (xyz) to the 2008th power? Requiring rigorous steps,

Given that the product of the x-th power of 2 by the y-th power of 3 by the z-th power of 111 is equal to 1998, where x.y.z is a natural number, how to find the value of (xyz) to the 2008th power? Requiring rigorous steps,

2,3 Are prime numbers,111 factorization:111=3*37.1998 factorization:1998=2*3*3*3*37, note that the index of 37 is 1. so z=1, x=1, y=2. The result is 2^2008.

Given that the X-th power of 2 is multiplied by the Y-th power of 3 and the Z-th power of 23=1242, where XYZ is an integer, find (X-Y+Z) of the 2010th power Given that the X-th power of 2 is multiplied by the Y-th power of 3 and the Z-th power of 23=1242, where XYZ is an integer, find the (X-Y+Z) of the 2010th power Given that the X-th power of 2 is multiplied by the Y-th power of 3 and the Z-th power of 23=1242, where XYZ is an integer, find the 2010th power of (X-Y+Z

Decompose 1242 into 2*3*3*3*23
So x=1 y=3 z=1
X-y+z=1-3+1=-1
2010 Power of -1=1