Question 1011452
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A) {{{sqrt(2)}}} . {{{root(3,16)}}} . {{{root(6,2)}}} = 2^(1/2 + 4/3 + 1/6) = {{{2^(1/2 + 4/3 + 1/6)}}} = {{{2^2}}} = 4  is an integer.  <-----  {{{1/2 + 4/3 + 1/6}}} = {{{3/6 + 8/6 + 1/6}}} = {{{(3 + 8 + 1)/6}}} = {{{12/6}}} = 2.
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