Answer
$\frac{2}{9}$
Work Step by Step
$(\frac{2}{3})^{2}.(\frac{1}{3}+\frac{1}{6})$
= $(\frac{2}{3}\times\frac{2}{3})\times(\frac{1}{3}+\frac{1}{6})$ (First solve bracket)
= $(\frac{2\times2}{3\times3})\times(\frac{1\times2}{3\times2}+\frac{1}{6})$
(least common multiple of 3 &6 is 6)
= $\frac{4}{9}\times(\frac{2}{6}+\frac{1}{6})$
= $\frac{4}{9}\times(\frac{2+1}{6})$ (add numerators)
= $\frac{4}{9}\times(\frac{3}{6})$
= $\frac{2\times2}{3\times3}\times(\frac{3}{2\times3})$
= $\frac{2}{9}$
