Đáp án:
$\left( {a;\,\,b} \right) = \left\{ {\left( {1;\,\,2} \right),\,\,\,\left( {2;\,\,1} \right)} \right\}.$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{a^3} + {b^3} = 9\\
a + b = 3 \Rightarrow b = 3 - a
\end{array} \right.\\
{a^3} + {b^3} = 9\\
\Leftrightarrow \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = 9\\
\Leftrightarrow 3\left( {{a^2} + 2ab + {b^2} - 3ab} \right) = 9\\
\Leftrightarrow {\left( {a + b} \right)^2} - 3ab = 3\\
\Leftrightarrow {3^2} - 3ab = 3\\
\Leftrightarrow 3ab = 6 \Leftrightarrow ab = 2\\
\Leftrightarrow a\left( {3 - a} \right) = 2\\
\Leftrightarrow 3a - {a^2} = 2\\
\Leftrightarrow {a^2} - 3a + 2 = 0\\
\Leftrightarrow a - 2a - a + 2 = 0\\
\Leftrightarrow a\left( {a - 2} \right) - \left( {a - 2} \right) = 0\\
\Leftrightarrow \left( {a - 1} \right)\left( {a - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a - 1 = 0\\
a - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 1 \Rightarrow b = 2\\
a = 2 \Rightarrow b = 1
\end{array} \right..
\end{array}$
