Question 737616


To find a counter example to "{{{(x+ y)^2}}} is {{{NOT}}} {{{x^2+ y^2}}}", you must use {{{x}}} and {{{y}}} so that {{{(x+y)^2= x^2+ y^2}}}.

Presumably you know that {{{(x+ y)^2= x^2+ 2xy+ y^2}}}. That means you must find {{{x}}} and {{{y}}} so that {{{x^2+ 2xy+ y^2= x^2+ y^2}}}.

{{{cross(x^2)+ 2xy+ cross(y^2)= cross(x^2)+ cross(y^2)}}}...notice that the squares on both sides cancel

 What does that leave you with? 

{{{2xy= 0}}}...this is true only if only {{{x=0}}} or {{{y=0}}}, or both {{{x=0}}} and {{{y=0}}}

so, it proves that {{{(x+ y)^2<>x^2+ y^2}}}