Answer
$\Big(\dfrac{2}{3}\Big)^{3}+\dfrac{1}{9}+\dfrac{1}{3}\cdot\dfrac{4}{3}=\dfrac{23}{27}$
Work Step by Step
$\Big(\dfrac{2}{3}\Big)^{3}+\dfrac{1}{9}+\dfrac{1}{3}\cdot\dfrac{4}{3}$
First, evaluate the exponential expression:
$\Big(\dfrac{2}{3}\Big)^{3}+\dfrac{1}{9}+\dfrac{1}{3}\cdot\dfrac{4}{3}=\dfrac{8}{27}+\dfrac{1}{9}+\dfrac{1}{3}\cdot\dfrac{4}{3}=...$
Now, evaluate the product:
$...=\dfrac{8}{27}+\dfrac{1}{9}+\dfrac{4}{9}=...$
Finally, evaluate the sums and simplify if possible:
$...=\dfrac{8}{27}+\dfrac{5}{9}=\dfrac{(8)(9)+(27)(5)}{(27)(9)}=...$
$...=\dfrac{72+135}{243}=\dfrac{207}{243}=\dfrac{23}{27}$
