解：原式?$=4a^2-b^2+3(4a^2-4ab+b^2)$?
?$=16a^2-12ab+2b^2$?
?$=4a(4a-3b)+2b^2$?
將?$a=\frac {3} {4}，$??$b=\frac {1} {3}$?代入得：
?$ 4a(4a-3b)+2b^2=4×\frac {3} {4}×(4×\frac {3} {4}-3×\frac {1} {3})+2×{(\frac {1} {3})}^2$?
?$ =3×(3-1)+\frac {2} {9}$?
?$ =\frac {56} {9}$?
