Question 277236
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What is the square root of 9y to the 6th power?
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@mananth in his post treats it as   {{{sqrt(9y^6)}}},  and gives the answer   {{{3y^3}}}.


@dabanfield in his post treats it as   {{{sqrt((9y)^6)}}},  and gives the answer   {{{729y^3}}}.



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                They both are incorrect.


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If to treat it as   {{{sqrt(9y^6)}}},  then the correct answer is    {{{3*abs(y)^3}}}   with the use of the absolute value sign.


If to treat it as   {{{sqrt((9y)^6)}}},  then the correct answer is    {{{729*abs(y)^3}}}   with the use of the absolute value sign.



The error of the two other persons is that value  ' y '  in the input formula can be negative.


Then both other persons return the negative value of the square root, which contradicts 
to the common agreement about the square root values.


My formula works universally for positive and negative values of  ' y ' - - -  it is why it is preferable
and why it is the uniquely correct form.



Solved.



The problems of this kind are a standard TRAP to catch unprepared/undertrained novices.