Đáp án:
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{2k + 2}}{k}\\
{x_1}{x_2} = \frac{{k + 2}}{k}
\end{array} \right.\\
a)x_1^3 - 3{x_1}{x_2} + x_2^3\\
= {\left( {{x_1} + {x_2}} \right)^3} - 3{x_1}{x_2}\left( {{x_1} + {x_2}} \right) - 3{x_1}{x_2}\\
= \frac{{{{\left( {2k + 2} \right)}^3}}}{{{k^3}}} - 3.\frac{{k + 2}}{k}.\frac{{2k + 2}}{k} - 3.\frac{{k + 2}}{k}\\
= \frac{{8{k^3} + 24{k^2} + 24k + 8 - 6k\left( {k + 2} \right)\left( {k + 1} \right) - 3{k^2}\left( {k + 2} \right)}}{{{k^3}}}\\
= \frac{{ - {k^3} + 18k + 8}}{{{k^3}}}\\
b)x_1^2 + x_2^2 = 1\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 1\\
\Rightarrow \frac{{{{\left( {2k + 2} \right)}^2}}}{{{k^2}}} - 2.\frac{{k + 2}}{k} = 1\\
\Rightarrow \frac{{4{k^2} + 8k + 4 - 2{k^2} - 4k}}{{{k^2}}} = 1\\
\Rightarrow 2{k^2} + 4k + 4 = {k^2}\\
\Rightarrow {k^2} + 4k + 4 = 0\\
\Rightarrow {\left( {k + 2} \right)^2} = 0\\
\Rightarrow k = - 2
\end{array}$
