Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_3} + {u_6} = 23\\
{u_3}^3 + {u_6}^3 = 4439
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_6} = 23 - {u_3}\\
{u_3}^3 + {\left( {23 - {u_3}} \right)^3} = 4439
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_6} = 23 - {u_3}\\
{u_3}^3 + {23^3} - {3.23^2}.{u_3} + 3.23.{u_3}^2 - {u_3}^3 = 4439
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_6} = 23 - {u_3}\\
69{u_3}^2 - 1587{u_3} + 7728 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{u_3} = 16 \Rightarrow {u_6} = 7\\
{u_3} = 7 \Rightarrow {u_3} = 16
\end{array} \right.\\
TH1:\,\,\,\left\{ \begin{array}{l}
{u_3} = 16\\
{u_6} = 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 2d = 16\\
{u_1} + 5d = 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 22\\
d = - 3
\end{array} \right.\\
{u_n} = {u_1} + \left( {n - 1} \right)d = 22 + \left( {n - 1} \right).\left( { - 3} \right) = 25 - 3n\\
TH2:\,\,\,\left\{ \begin{array}{l}
{u_3} = 7\\
{u_6} = 16
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 2d = 7\\
{u_1} + 5d = 16
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 1\\
d = 3
\end{array} \right.\\
{u_n} = {u_1} + \left( {n - 1} \right)d = 1 + \left( {n - 1} \right).3 = 3n - 2
\end{array}\)
