is there are any short cut method to find the determinant of A? Find the determinant  of A

$$A=\left(\begin{matrix} 
  x^1 & x^2 & x^3 \\
  x^8 & x^9 & x^4 \\
  x^7 & x^6 & x^5 \\
\end{matrix}\right)$$

My attempts : By  doing a laplace expansion along the first  column  i can calculate,but it is  a long process My question is that  is there  are  any short cut method  to find the determinant of A?
Thanks u
 A: \begin{align*}
\det(A) =\begin{vmatrix} 
  x^1 & x^2 & x^3 \\
  x^8 & x^9 & x^4 \\
  x^7 & x^6 & x^5 \\
\end{vmatrix}
&=\color{red}{x^5}\begin{vmatrix} 
  x^1 & x^2 & x^3 \\
  x^8 & x^9 & x^4 \\
  \color{red}{x^2} & \color{red}{x^1} & \color{red}{1} 
\end{vmatrix}\\
&=x^5\cdot x^4 \cdot x^1\begin{vmatrix} 
  1 & x & x^2 \\
  x^4 & x^5 & 1 \\
  x^2 & x & 1 \\
\end{vmatrix}\\
&=x^{10}\cdot \color{red}{x^1}\begin{vmatrix} 
  \color{blue}{1} & \color{red}{1} & x^2 \\
  \color{blue}{x^4} & \color{red}{x^4} & 1 \\
  \color{blue}{x^2} & \color{red}{1} & 1 
\end{vmatrix}\\
&=x^{11}\begin{vmatrix} 
  \color{blue}{0} & 1 & x^2 \\
  \color{blue}{0} & x^4 & 1 \\
  \color{blue}{x^2-1} & 1 & 1 
\end{vmatrix}\\
&=x^{11}(x^2-1)(1-x^6).
\end{align*}
A: Use Gaussian elimination, while keeping track of changes to the determinant.
$$ \begin{aligned}
\left\lvert \begin{matrix} x^1 & x^2 & x^3 \\ x^8 & x^9 & x^4 \\ x^7 & x^6 & x^5 \end{matrix} \right\rvert
&= x^6 \left\lvert \begin{matrix} 1 & 1 & 1 \\ x^7 & x^7 & x^1 \\ x^6 & x^4 & x^2 \end{matrix} \right\rvert \\
&= x^6\left\lvert \begin{matrix} 1 & 0 & 0 \\ x^7 & 0 & x^1 - x^7 \\ x^6 & x^4 - x^6 & x^2 - x^6 \end{matrix} \right\rvert \\
&= x^6\left\lvert \begin{matrix}0 & x^1 - x^7 \\ x^4 - x^6 & x^2 - x^6 \end{matrix} \right\rvert \\
&= -x^{11}(1 - x^6)(1 - x^2)
\end{aligned}$$
A: Do you know the "spaghetti rule" for calculating $3\times 3$ determinants?$$
\det\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\end{bmatrix}= 
a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{21}a_{32}a_{13}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}$$
This is actually quite easy to remember.  The terms with plus signs are the products of the elements on the broken diagonals parallel to the main diagonal, and those with minus signs are the products of the elements on the broken diagonals parallel to the minor diagonal.
Knowing this rule, you can just write down the answer to your problem immediately.
