Cho hệ phương trình: $\left\{ \begin{array}{l}\left( {a + b} \right)x + \left( {a - b} \right)y = 2\\\left( {{a^3} + {b^3}} \right)x + \left( {{a^3} - {b^3}} \right)y = 2\left( {{a^2} + {b^2}} \right)\end{array} \right.$. Với $a \ne  \pm b;a,b \ne 0$, hệ phương trình có nghiệm duy nhất bằng

Ta có :

$\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}{a + b}&{a - b}\\{{a^3} + {b^3}}&{{a^3} - {b^3}}\end{array}} \right| = \left( {a + b} \right)\left( {{a^3} - {b^3}} \right) - \left( {a - b} \right)\left( {{a^3} + {b^3}} \right)\\\,\,\,\,\, = \left( {a + b} \right)\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) - \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\\,\,\,\,\, = \left( {a + b} \right)\left( {a - b} \right)\left( {{a^2} + ab + {b^2} - {a^2} + ab - {b^2}} \right) = 2{\rm{a}}b\left( {a + b} \right)\left( {a - b} \right)\\{D_x} = \left| {\begin{array}{*{20}{c}}2&{a - b}\\{2({a^2} + {b^2})}&{{a^3} - {b^3}}\end{array}} \right| = 2\left( {{a^3} - {b^3}} \right) - 2\left( {a - b} \right)\left( {{a^2} + {b^2}} \right)\\\,\,\,\,\,\,\, = 2\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) - 2\left( {a - b} \right)\left( {{a^2} + {b^2}} \right)\\\,\,\,\,\,\,\, = 2ab\left( {a - b} \right)\\{D_y} = \left| {\begin{array}{*{20}{c}}{a + b}&2\\{{a^3} + {b^3}}&{2\left( {{a^2} + {b^2}} \right)}\end{array}} \right| = 2\left( {a + b} \right)\left( {{a^2} + {b^2}} \right) - 2\left( {{a^3} + {b^3}} \right)\\\,\,\,\,\,\,\, = 2\left( {a + b} \right)\left( {{a^2} + {b^2}} \right) - 2\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\\,\,\,\,\,\,\, = 2ab\left( {a + b} \right)\end{array}$

Với $a \ne  \pm b;a,b \ne 0 \Rightarrow D \ne 0$, hệ phương trình có nghiệm duy nhất $\left\{ \begin{array}{l}x = \dfrac{{{D_x}}}{D} = \dfrac{{2ab(a - b)}}{{2ab(a - b)(a + b)}} = \dfrac{1}{{a + b}}\\y = \dfrac{{{D_y}}}{D} = \dfrac{{2ab(a + b)}}{{2ab(a - b)(a + b)}} = \dfrac{1}{{a - b}}\end{array} \right.$