原式=4a2-b2+3(4a2+b2-4ab)-12a2+9ab+6a2-3ab-2ab+b2
=4a2-b2+12a2+3b2-12ab-12a2+9ab+6a2-3ab-2ab+b2
=10a2+3b2-8ab,
当a=-1,b=2时,原式=10×1+3×4-8×(-1)×2=10+12+8=30.
解答:解:3(2a-b)2+(-3a)(4a-3b)
=12a2-12ab+3b2-12a2+9ab
=-3ab+3b2,
当a=-1,b=-2时,原式=-3×(-1)×(-2)+3×(-2)2=6.