A better way to evaluate a certain determinant 
Question Statement:-
Evaluate the determinant:
  $$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\ 
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$


My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\ 
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &= 
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 1 & 1 \\ 
2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\
3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\
C_2\rightarrow\dfrac{C_2}{2^2}\\
C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\
&=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 0 & 0 \\ 
2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\
3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\
\end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\
C_3\rightarrow C_3-C_1\end{array}\right]\\
&=
(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\ 
2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\
3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}\\
&=(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\ 
2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\
3^2 & 5 & \dfrac{56}{9} \\
\end{vmatrix}\\
&=(1^2\times2^2)
\begin{vmatrix}
1 & 0 & 0 \\ 
2^2 & \dfrac{7}{4} & 20 \\
3^2 & 5 & 56 \\
\end{vmatrix}\\
&=4\times(-2)\\
&=-8
\end{align}
$$
As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.

Using the rule of Sarrus, the computation is really not too long, and we get in general for all $n\ge 1$,
$$
\det \begin{pmatrix} n^2 & (n+1)^2 & (n+2)^2\cr (n+1)^2& (n+2)^2 & (n+3)^2\cr
(n+2)^2& (n+3)^2 & (n+4)^2\end{pmatrix}=-8.
$$
