Ans: (a) $b(3-a)$ (b) $(3+a)(3-a)$ (c) $(3-a)(3+a+b)$
-
$\begin{array}{cl}
& 3b-ab \\
= & b(3-a)
\end{array}$ -
$\begin{array}{cl}
& 9-a^2 \\
= & (3+a)(3-a)
\end{array}$ -
$\begin{array}{cl}
& 9-a^2 + 3b – ab \\
= & (3+a)(3-a) + b(3-a) \\
= & (3-a)(3+a+b) \\
\end{array}$
