Coefficient of Power of x Using Difference of Squares With Unmatched Powers

To find the coefficient of  
\[x^4\]
  in the expansion of  
\[(3-x)^6(3+x)^4\]
  notice that  
\[9-x^2=(3-x)(3+x)\]
  so
Write the question as  
\[(3-x)^2(3-x)^4(3+x)^4=(9-6x+x^2)(9-x^2)\]
.
We are only interested in the coefficient of  
\[x^4\]
  so ignore any powers of  
\[x\]
  higher than  
\[x^4\]
.
\[\begin{equation} \begin{aligned} (3-x)^6(3+x)^4 &= (9-6x_x^2)(9-x^2)^4 \\ &= (9-6x+x^2)({}^4C_09^4(-x^2)^0+{}^4C_1 9^3(-x^2)^1+{}^4C_2 9^2(-x^2)^2+ higher \; powers \; of \; x) \\ &= (9-6x+x^2)(6561-2916x^2+486x^4 + higher \; powers \; of \; x ) \end{aligned} \end{equation}\]

The only contributions to the coefficient of  
\[x^4\]
  are from  
\[x^2 \times -2916 x^2\]
  and  
\[9 \times 486x^4=4374x^4\]
. The coefficient is  
\[-2916+4374=1458\]
.

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