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

\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*}
