Question 75102
{{{(81x^12y^8z^16)^(1/4)}}}
{{{3x^3y^2z^4}}}
because
{{{3^4=81}}}
{{{(x^3)^4=x^12}}}
{{{(y^2)^4=y^8}}}
{{{(z^4)^4=z^16}}}