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.

\begin{align}
   \begin{vmatrix}
      x^2     & y^2     & z^2 \\
      (x+1)^2 & (y+1)^2 & (z+1)^2 \\
      (x+2)^2 & (y+2)^2 & (z+2)^2 \\
   \end{vmatrix} &=
   \begin{vmatrix}
      x^2  & y^2  & z^2  \\
      2x+1 & 2y+1 & 2z+1 \\
      4x+4 & 4y+4 & 4z+4 \\
   \end{vmatrix}
\\ &=
   \begin{vmatrix} 
      x^2  & y^2  & z^2  \\
      2x+1 & 2y+1 & 2z+1 \\
      2    & 2    & 2    \\
   \end{vmatrix}
\\ &=
   \begin{vmatrix} 
      x^2-z^2 & y^2-z^2  & z^2  \\
      2x-2z   & 2y-2z    & 2z+1 \\
      0       & 0        & 2    \\
   \end{vmatrix}
\\ &= 4(x^2-z^2)(y-z) - 4(y^2-z^2)(y-z)
\\ &= 4(x-z)(y-z)[(x+z) - (y+z)]
\\ &= 4(x-z)(y-z)(x-y)]
\\ &= -4(x-y)(y-z)(z-x)]
\end{align}
When $x=n, \quad y=n+1, \quad z = n+2$, then
$\begin{vmatrix}
      n^2     & (n+1)^2 & (n+2)^2 \\
      (n+1)^2 & (n+2)^2 & (n+3)^2 \\
      (n+2)^2 & (n+3)^2 & (n+4)^2 \\
\end{vmatrix}
=-4(-1)(-1)(2) = -8$
