(x - 2·b)·(b - x) - x·(3·b - x) - (- 2·x)^2=
=(- x^2 + 3·b·x - 2·b^2) - (3·b·x - x^2) - 4·x^2=
=- 4·x^2 - 2·b^2
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$\small (x-2b)(b-x)-x(3b-x)-(-2x)^2 =$
$\small = bx-\cancel{x^2}-2b^2+2bx-3bx+\cancel{x^2}-4x^2 =$
$\small = (1+2-3)bx-2b^2-4x^2 =$
$\small = 0bx-2b^2-4x^2 =$
$\small = -2b^2-4x^2 $