## Intermediate Algebra (12th Edition)

The solution below shows that $(x+y)^2$ is not equal to $x^2+y^2.$
Substituting $x=3$ and $y=4,$ the given expression, $(x+y)^2 ,$ evaluates to \begin{array}{l}\require{cancel} (3+4)^2 \\\\= (7)^2 \\\\= 49 ,\end{array} while the given expression, $x^2+y^2 ,$ evaluates to \begin{array}{l}\require{cancel} 3^2+4^2 \\\\= 9+16 \\\\= 25 .\end{array}