System of equations
$(k+1)^3 x+(k+2)^3 y=(k+3)^3$
$(k+1) x+(k+2) y=(k+3)$
$x+y=1$ is consistent.
Since, the given system of equations are consistent. Then $D=0$ and Also, $\left(D_1=D_2=D_3=0\right)$ have infinitely many solutions. By Crammer Rule
$D=\left|\begin{array}{ccc}(k+1)^3 & (k+2)^3 & (k+3)^3 \\(k+1) & (k+2) & (k+3) \\1 & 1 & 1\end{array}\right|=0$
$\because \quad C_2 \rightarrow C_2-C_1$
$C_3 \rightarrow C_3-C_1$
$D=\left|\begin{array}{ccc}(k+1)^3 & (k+2)^3-(k+1)^3 & (k+3)^3-(k+1)^3 \\ (k+1) & (k+2)-(k+1) & (k+3)-(k+1) \\ 1 & 0 & 0\end{array}\right|$
$=0$
$D=\left|\begin{array}{ccc}(k+1)^3 & \left(3 k^2+9 k+7\right) & \left(6 k^2+24 k+26\right) \\ (k+1) & 1 & 2 \\ 1 & 0 & 0\end{array}\right|$
$=0$
Expand with $r$ to $R_3$
$2\left(3 k^2+9 k+7\right)-2\left(3 k^2+12 k+13\right)=0$
$\Rightarrow 3 k^2+9 k+7-3 k^2-12 k-13=0$
$\Rightarrow \quad-3 k-6=0 \Rightarrow 3 k=-6$
$\Rightarrow \quad k=-2$