Answer
$-9a^2-b^2+6ab+16$
Work Step by Step
RECALL:
(i) $(x+y)(x-y) = x^2-y^2$
(ii) $(x-y)^2=x^2-2xy+y^2$
Use rule (i) above where $x=4$ and $y=3a-b$ to have:
$\\=4^2-(3a-b)^2$
Use rule (ii) above where $x=3a$ and $y=b$ to have:
$\\=16-[(3a)^2-2(3a)(b)+b^2]
\\=16-(9a^2-6ab+b^2)
\\=16-9a^2-(-6ab)-b^2
\\=16-9a^2+6ab-b^2
\\=-9a^2-b^2+6ab+16$
