Question 170952
Note: {{{(x^y)^z=x^(y*z)}}}. Notice how the exponents are being multiplied.


Also, we can write 3 as {{{3^1}}}, 2 as {{{2^1}}}, and "z" as {{{z^1}}} which will make the expression {{{((3^1y^8)/(2^1z^1y^2))^4}}}



{{{((3^1y^8)/(2^1z^1y^2))^4}}} Start with the given expression



{{{(3^1y^8)^4/(2^1z^1y^2)^4}}} Break up the fraction



{{{(3^(1*4)y^(8*4))/(2^(1*4)z^(1*4)y^(2*4))}}} Multiply the outer exponent by EVERY exponent inside the parenthesis



{{{(3^(4)y^(32))/(2^(4)z^(4)y^(8))}}} Multiply the exponents.



{{{(81y^(32))/(16z^(4)y^(8))}}} Raise 3 to the 4th power to get 81. Raise 2 to the 4th power to get 16. 



{{{(81y^(32-8))/(16z^(4))}}} Now divide the "y" terms by subtracting their exponents



{{{(81y^(24))/(16z^(4))}}} Subtract



So {{{((3y^8)/(2zy^2))^4}}} simplifies to {{{(81y^(24))/(16z^(4))}}}