Prove that `t^(2) + 3t + 3` is a factor of `( t+1)^(n+1) + (t+2)^(2n -1)` for all intergral values of n `in` N.

Explanation

Put `t + 1= Z`.
Then `t^(2) + 3t + 3=(t+1)^(2) +(t+2)=Z^(2) + Z +1 = (Z-omega) (Z-omega)`
where `omega` is a complex cube root of uninty.
When `Z= omega`, the experssion `(t+1)^(n+1) + (t+2)^(2n-1)` becomes
`omega^(n+1)+(omega +1)^(2n-1) = omega^(n+1)+(t+2)^(2n+1)` becomes
`omega^(n+1) +(omega +1)^(2n-1) = omega^(n+1) +(-omega^(2))^(2n-1)`
`= omega^(n+1) +(-1)^(2n-1) omega^(4n-2)`
`= omega^(n+1) {1+(-1)^(2n-1) omega^(3n-3)}`
`= omega^(n+1) (1-1) = 0 as omega^(3n-3)= 1`
`threfore Z - omega` is a factor of the experssion .
Similary, `Z - omega^(2)` is also a factors of the expression.