Question 947109
{{{HM=2/(1/(3+sqrt(27))+1/(3-3sqrt(3)))}}}
{{{HM=2/(1/(3+3sqrt(3))+1/(3-3sqrt(3)))}}}
Let's work on the denominator,
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=((3-3sqrt(3)))/((3+3sqrt(3))(3-3sqrt(3)))+((3+3sqrt(3)))/((3+3sqrt(3))(3-3sqrt(3)))}}}
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=(3-3sqrt(3)+3+3sqrt(3))/((3+3sqrt(3))(3-3sqrt(3)))}}}
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=(6)/((3+3sqrt(3))(3-3sqrt(3)))}}}
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=(6)/(9-9sqrt(3)+9sqrt(3)-27)}}}
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=(6)/(-18)}}}
{{{(1/(3+3sqrt(3))+1/(3-3sqrt(3)))=(1)/(-3)}}}
Now substituting into the denominator,
{{{HM=2/(-(1/3))}}}
{{{HM=-6}}}