We have, f(n)=A(−2)n+B(−3)n f(n)+af(n−1)+bf(n−2)=0 ∴‌‌A(−2)n+B(−3)n+a(A(−2)n−1+B(−3)n−1) +b(A(−2)n−2+B(−3)n−2=0 ⇒A(−2)n−2[4−2a+b]+(−3)n−2B[9−3a+b]=0 ∴ It is possible only 4−2a+b=0‌‌ and9−3a+b=0 Solving, we get a=5,b=6 ∴(a+b)(b−a)=(5+6)(6−5)=11