Sol. Put t + 1 = Z. Then t 2 +3t + 3 = (t + 1) 2 + (t + 2) = Z 2 + Z+ 1 and so (Z 2 + Z + 1) = (Z – ω ) (Z – ω 2 ), where ω is a complex cube root of unity. When Z = ω , the expression (t + 1) n+1 + ( t + 2) 2n–1 becomes

ω n+1 + ( ω + 1) 2n–1 = ω n+1 + (– ω 2 ) 2n–1

= ω n+1 + (–1) 2n–1 ω 4n–2

= ω n+1 {1 +(–1) 2n–1 ω 3n–3 }

= ω n+1 (1–1) = 0 as ω 3n –3 = 1

∴ Z – ω is a factor of the expression.

Similarly, Z – ω 2 is also a factor of the expression,

∴ Z 2 + Z + 1 is a factor of (t + 1) n+1 + (t + 2) 2n–1

i.e. t 2 + 3t + 3 is a factor of (t + 1) n+1 + (t + 2) 2n–1