Question 1116399
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The way  (= <U>the ONLY way</U>;  == <U>the CANONICAL way</U>) of solving this problem is  <U>THIS</U>:



1.  1 = {{{(a+b)^3}}} = {{{a^3 + 3a^2*b + 3a*b^2 + b^3}}} = {{{(a^3+b^3)}}} + {{{3ab(a+b)}}} = (replace a^3 + b^3 by 16 and replace a+b by 1, since it is given) = 


        16 + 3ab,

    which implies  3ab = 1 - 16 = -15  and hence  ab = -5.    (*)



2.  Now the next and the last step is  


    1 = {{{(a+b)^2}}} = {{{a^2 + 2ab + b^2}}} = {{{(a^2+b^2)}}} + {{{(2ab)}}} = (replace ab by -5, since we just found it in (*)) = {{{a^2 + b^2}}} - {{{2*5}}} = {{{a^2 + b^2}}} - 10,


    which implies  {{{a^2 + b^2}}} = 1 + 10 = 11.


<U>Answer</U>. Under given conditions,  {{{a^2 + b^2}}} = 11.
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