(2b-3c+4)(3c-2b+4)-2(b-c)(b-c)
=[4+(2b-3c)][4-(2b-3c)]-2(b-c)^2
=4^2-(2b-3c)^2-4(b^2-2bc+c^2)
=16-(4b^2-12bc+9c^2)-4b^2+8bc-4c^2
=16-4b^2+12bc-9c^2-4b^2+8bc-4c^2
=16-8b^2+20bc-13c^2.
(2b-3c+4)(3c-2b+4)-2(b-c)(b-c)
=[4+(2b-3c)][4-(2b-3c)]-2(b-c)^2
=4^2-(2b-3c)^2-4(b^2-2bc+c^2)
=16-(4b^2-12bc+9c^2)-4b^2+8bc-4c^2
=16-4b^2+12bc-9c^2-4b^2+8bc-4c^2
=16-8b^2+20bc-13c^2.