Question 975329
{{{(x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+3xy(x+y)+y^3}}}
When {{{x+y=2}}} ,
{{{(x+y)^3=2^3=8}}} and {{{(x+y)^3=x^3+3xy(x+y)+y^3=x^3+3xy(2)+y^3=x^3+6xy+y^3}}} ,
so {{{x^3+6xy+y^3=8}}}--->{{{x^3+6xy+y^3-8=0}}}