By using properties of determinants,show that:
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3y+k)$

(A) Let $\Delta = \left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$.
Applying the row operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$,we get:
$\Delta = \left|\begin{array}{ccc}3y+k & 3y+k & 3y+k \\ y & y+k & y \\ y & y & y+k\end{array}\right|$.
Taking $(3y+k)$ common from $R_{1}$:
$\Delta = (3y+k) \left|\begin{array}{ccc}1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k\end{array}\right|$.
Applying column operations $C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}$:
$\Delta = (3y+k) \left|\begin{array}{ccc}1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k\end{array}\right|$.
Expanding along the first row:
$\Delta = (3y+k) \cdot 1 \cdot (k \cdot k - 0 \cdot 0) = k^{2}(3y+k)$.
Thus,the result is proved.