典例3 分解因式：
(1)$25(x-2y)^2 - 4(y-2x)^2$；
(2)$2(x^2 - \frac{1}{2}) - x^4$。

解 (1)$25(x-2y)^2 - 4(y-2x)^2$
$=[5(x-2y)+2(y-2x)][5(x-2y)-2(y-2x)]$
$=(5x -10y +2y -4x)(5x -10y -2y +4x)$
$=(x -8y)(9x -12y)$
$=3(x -8y)(3x -4y).$
(2)$2(x^2 - \frac{1}{2}) - x^4$
$=2x^2 -1 -x^4$
$=-(x^4 -2x^2 +1)$
$=-(x^2 -1)^2$
$=-(x+1)^2(x-1)^2.$

典例2
解 $6x(x-y)^2 + 3(y-x)^3$
$=6x(x-y)^2 - 3(x-y)^3$
$=3(x-y)^2[2x-(x-y)]$
$=3(x-y)^2(x+y).$

解：
$\begin{aligned}6x(x-y)^2 + 3(y-x)^3&=6x(x-y)^2 - 3(x-y)^3\\&=3(x-y)^2[2x - (x-y)]\\&=3(x-y)^2(x + y)\end{aligned}$

解 (1)$25(x-2y)^2 - 4(y-2x)^2$
$=[5(x-2y)+2(y-2x)][5(x-2y)-2(y-2x)]$
$=(5x -10y +2y -4x)(5x -10y -2y +4x)$
$=(x -8y)(9x -12y)$
$=3(x -8y)(3x -4y).$
(2)$2(x^2 - \frac{1}{2}) - x^4$
$=2x^2 -1 -x^4$
$=-(x^4 -2x^2 +1)$
$=-(x^2 -1)^2$
$=-(x+1)^2(x-1)^2.$

解：
(1)
$\begin{aligned}25(x-2y)^2 - 4(y-2x)^2&=[5(x-2y)+2(y-2x)][5(x-2y)-2(y-2x)]\\&=(5x-10y+2y-4x)(5x-10y-2y+4x)\\&=(x-8y)(9x-12y)\\&=3(x-8y)(3x-4y)\end{aligned}$
(2)
$\begin{aligned}2(x^2-\frac{1}{2})-x^4&=2x^2-1-x^4\\&=-(x^4-2x^2+1)\\&=-(x^2-1)^2\\&=-(x+1)^2(x-1)^2\end{aligned}$