## 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{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}

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
$-2916+4374=1458$
.