$\begin{array}{l} A = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) - \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\ A = {x^3} - {y^3} - \left( {{x^3} + {y^3}} \right) = - 2{y^3}\\ C = \left( {2x + 3y} \right)\left( {4{x^2} - 6xy + 9{y^2}} \right)\\ C = {\left( {2x} \right)^3} + {\left( {3y} \right)^3} = 8{x^3} + 27{y^3}\\ 2)\\ A = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\ A = \left( {{x^3} + 1} \right) - \left( {{x^3} - 1} \right) = {x^3} - {x^3} + 1 + 1 = 2\\ 3)\\ b)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - x\left( {x - 4} \right)\left( {x + 4} \right) = 21\\ \Leftrightarrow {x^3} - 27 - x\left( {{x^2} - 16} \right) = 21\\ \Leftrightarrow {x^3} - 27 - {x^3} + 16x = 21\\ \Leftrightarrow 16x = 48\\ \Leftrightarrow x = 3\\ 4)\\ {x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\ = 1\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right] = 1\left[ {1 - 3\left( { - 1} \right)} \right] = \left( {1 + 3} \right) = 4\\ 5)\\ A = \dfrac{{{{2020}^3} + 1}}{{{{2020}^2} - 2019}}\\ A = \dfrac{{\left( {2020 + 1} \right)\left( {{{2020}^2} - 2020 + 1} \right)}}{{{{2020}^2} - 2019}} = \dfrac{{2021\left( {{{2020}^2} - 2019} \right)}}{{{{2020}^2} - 2019}}\\ = 2021\\ b)B = \dfrac{{{{2020}^3} - 1}}{{{{2020}^2} + 2021}} = \dfrac{{\left( {2020 - 1} \right)\left( {{{2020}^2} + 2020 + {1^2}} \right)}}{{{{2020}^2} + 2021}}\\ B = 2019 \end{array}$
