[tex]\left(1+\dfrac{1}{3}\right)^{2}-\dfrac{2}{9}=1+\dfrac{2}{3}+\dfrac{1}{9}-\dfrac{2}{9}\\\\=1+\dfrac{6+1-2}{9}=1\frac{5}{9}[/tex]
Another way to compute this is ...
[tex]\left(1+\dfrac{1}{3}\right)^{2}-\dfrac{2}{9}=\left(\dfrac{4}{3}\right)^{2}-\dfrac{2}{9}=\dfrac{16}{9}-\dfrac{2}{9}\\\\=\dfrac{16-2}{9}=\dfrac{14}{9}=1\frac{5}{9}[/tex]
