Question 206178
{{{(16x^15y^7z^11)^(1/3)}}}
={{{(2^4*x^15*y^7*z^11)^(1/3)}}}
Pull out the greatest factor of each that has an exponent divisible by 3.
={{{(2^3*2^1*x^15*y^6*y^1*z^9*z^2)^(1/3)}}}
For each factor that has an exponent divisible by 3, there is a perfect cube root of that base with the exponent divided by 3.
={{{2*x^5*y^2*z^3*(2*y*z^2)^(1/3))}}}