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.
 A: The direct formula for $3$ by $3$ determinants isn't so bad
$$\begin{vmatrix}
a & b & c \\ 
d & e & f \\
g & h & i \\
\end{vmatrix}=aei+bfg+cdh-ceg-bdi-afh$$
so
$$\begin{vmatrix}
1 & 4 & 9 \\ 
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=225+576+576-729-400-256=-8.$$
Row operations and other similar tricks tend to speed things up only when the matrix is $4$ by $4$ or larger.
A: If we consider
$$ f(x)=\det\begin{pmatrix}x^2 & (x+1)^2 & (x+2)^2 \\ (x+1)^2 & (x+2)^2 & (x+3)^2 \\ (x+2)^2 & (x+3)^2 & (x+4)^2 \end{pmatrix}$$
we have:
$$\scriptstyle f'(x) = \det\begin{pmatrix}2x & (x+1)^2 & (x+2)^2 \\ 2x+2 & (x+2)^2 & (x+3)^2 \\ 2x+4 & (x+3)^2 & (x+4)^2 \end{pmatrix}+\det\begin{pmatrix}x^2 & 2x+2 & (x+2)^2 \\ (x+1)^2 & 2x+4 & (x+3)^2 \\ (x+2)^2 & 2x+6 & (x+4)^2 \end{pmatrix}+\det\begin{pmatrix}x^2 & (x+1)^2 & 2x+4 \\ (x+1)^2 & (x+2)^2 & 2x+6 \\ (x+2)^2 & (x+3)^2 & 2x+8 \end{pmatrix} $$
and $f'(x)=0$ by Gaussian elimination. It follows that $f(x)$ is a constant function and
$$ f(1)=f(-2)=\det\begin{pmatrix}4 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 4\end{pmatrix}=-4-4=\color{red}{-8}.$$
A: Subtract the first column from other columns to reduce $n^2$-s, then subtract twice the second column from the third one to reduce $n$:
$$\begin{vmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\ 
(n+1)^2 & (n+2)^2 & (n+3)^2 \\
(n+2)^2 & (n+3)^2 & (n+4)^2
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 4n+4 \\ 
(n+1)^2 & 2n+3 & 4n+8 \\
(n+2)^2 & 2n+5 & 4n+12
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\ 
(n+1)^2 & 2n+3 & 2 \\
(n+2)^2 & 2n+5 & 2
\end{vmatrix}
$$
Next subtract the first row from other rows, then twice the second row from the third one:
$$\cdots
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\ 
2n+1 & 2 & 0 \\
4n+4 & 4 & 0
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\ 
2n+1 & 2 & 0 \\
2 & 0 & 0
\end{vmatrix}
$$
Now we have a triangular determinant with three twos on its antidiagonal, so the determinant is
$$2\cdot (-2)\cdot 2 = -8$$
A: Create two zeroes in the first row ($C_2 \to C_2-\color{red}{4}C_1$ and $C_3 \to C_3-\color{blue}{9}C_1$) and expand:
$$\begin{vmatrix}
1 & \color{red}{4} & \color{blue}{9} \\ 
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=
\begin{vmatrix}
1 & 0 & 0 \\ 
4 & 9-16 & 16-36 \\
9 & 16-36 & 25-81 \\
\end{vmatrix} = \begin{vmatrix}
-7 & -20 \\ 
-20 & -56 \\
\end{vmatrix} = 7 \cdot 56 - 20^2 = -8$$
A: Using the rule of Sarrus, the computation is really not too long, and we get in general for all $n\ge 1$,
$$
\det \begin{pmatrix} n^2 & (n+1)^2 & (n+2)^2\cr (n+1)^2& (n+2)^2 & (n+3)^2\cr
(n+2)^2& (n+3)^2 & (n+4)^2\end{pmatrix}=-8.
$$
A: \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
$\begin{vmatrix}
      n^2     & (n+1)^2 & (n+2)^2 \\
      (n+1)^2 & (n+2)^2 & (n+3)^2 \\
      (n+2)^2 & (n+3)^2 & (n+4)^2 \\
\end{vmatrix}
=-4(-1)(-1)(2) = -8$
