Factorise: $\dfrac{{8{a^3}}}{{27}} - \dfrac{{{b^3}}}{8}$

Hint:
It is given in the question that we have to Factorise $\dfrac{{8{a^3}}}{{27}} - \dfrac{{{b^3}}}{8}$
Firstly, we will write $\dfrac{{8{a^3}}}{{27}}$ as ${\left( {\dfrac{{2a}}{3}} \right)^3}$ and $\dfrac{{{b^3}}}{8}$ as ${\left( {\dfrac{b}{2}} \right)^3}$
After that, by applying formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ we will get our answer.

Complete step by step solution:
It is given in the question that we have to Factorise $\dfrac{{8{a^3}}}{{27}} - \dfrac{{{b^3}}}{8}$
 $ = \dfrac{{8{a^3}}}{{27}} - \dfrac{{{b^3}}}{8}$
We can write $\dfrac{{8{a^3}}}{{27}}$ as ${\left( {\dfrac{{2a}}{3}} \right)^3}$ and $\dfrac{{{b^3}}}{8}$ as ${\left( {\dfrac{b}{2}} \right)^3}$
Now, put this in the above equation.
  $ = {\left( {\dfrac{{2a}}{3}} \right)^3} - {\left( {\dfrac{b}{2}} \right)^3}$
Now, using formula of $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ in above equation we get,
 $ = \left( {\dfrac{{2a}}{3} - \dfrac{b}{2}} \right)\left[ {{{\left( {\dfrac{{2a}}{3}} \right)}^2} + \left( {\dfrac{{2a}}{3}} \right) \times \left( {\dfrac{b}{2}} \right) + {{\left( {\dfrac{b}{2}} \right)}^2}} \right]$
 $ = \left( {\dfrac{{2a}}{3} - \dfrac{b}{2}} \right)\left[ {\dfrac{{4{a^2}}}{9} + \dfrac{{ab}}{3} + \dfrac{{{b^2}}}{4}} \right]$

Note:
Some additional properties:
1) $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
2) ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
3) ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$