$ a)\left( {\frac{1}{3} + 2x} \right)\left( {4{x^2} - \frac{2}{3}x + \frac{1}{9}} \right) - \left( {8{x^3} - \frac{1}{{27}}} \right)\\$= $= \left( {2x + \frac{1}{3}} \right)\left( {{{\left( {2x} \right)}^2} - 2x.\frac{1}{3} + {{\left( {\frac{1}{3}} \right)}^2}} \right) - \left( {8{x^3} - \frac{1}{{27}}} \right)\\$ $= {\left( {2x} \right)^3} - {\left( {\frac{1}{3}} \right)^3} - \left( {8{x^3} - \frac{1}{{27}}} \right)\\$ $= 8{x^3} - \frac{1}{{27}} - \left( {8{x^3} - \frac{1}{{27}}} \right)\\$ $= 0\\$
$b){\left( {x - 1} \right)^3} - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 3\left( {1 - x} \right)x\\$ $= {\left( {x - 1} \right)^3} - \left( {{x^3} - 1} \right) - 3x + 3{x^2}\\ $=${\left( {x - 1} \right)^3} - \left( {{x^3} - 3{x^2} + 3x - 1} \right)\\$ $= {\left( {x - 1} \right)^3} - {\left( {x - 1} \right)^3}\\$ $= 0$
