Question 1109050
a). {{{(a+b)^3=a^3+3a^2b+3ab^2+b^3}}}
 
{{{a+b=1}}} --> {{{system((a+b)^3=1,"and", (a+b)^2=1)}}}
 
Substituting into {{{(a+b)^3=a^3+3a^2b+3ab^2+b^3}}} , we get
{{{1=16+3a^2b+3ab^2}}}
{{{-15=3a^2b+3ab^2}}}
{{{-15=3ab(a+b)}}}
{{{-15=3ab}}}
{{{-15/3=ab}}}
{{{ab=-5}}}
 
{{{a^2+b^2+2ab=(a+b)^2}}}
Substituting {{{(a+b)^2=1)}}} and {{{ab=-5}}} , we get
{{{a^2+b^2+2(-5)=1}}}
{{{a^2+b^2-10=1}}}
{{{a^2+b^2=1+10}}}
{{{a^2+b^2=11}}}
 
NOTE: You could use {{{system(a+b=1,ab=-5)}}} to find 
that {{{a}}} and {{{b}}} are {{{(1 +- sqrt(21))/2}}} ,
and then calculate {{{a^2+b^2}}} ,
but it is not fun.