If $$u_n = cos^n α + sin^n α$$, then the value of $$2u_6 - 3u_4 +1$$ is :

$$u_6 = cos^6 α + sin^6 α$$
$$2 u_6= 2cos^6 α + 2sin^6 α$$
$$u_4 = cos^4 α + sin^4 α$$
$$3 u_4= 3cos^4 α + 3sin^4 α$$
$$2 u_6 - 3 u_4 = (2cos^6 α + 2sin^6 α) - (3cos^4 α + 3sin^4 α)$$
$$2 cos^6 α = 2(1- sin^2 α)^3 = 2(1 - sin^3α- 3sin^2α+ 3 sin^4 α )$$
$$3 cos^4 α = 3(1- sin^2 α)^2 = 3(1 + sin^4α- 2sin^2α)$$
$$2 cos^6 α - 3 cos^4 α = 2(1 - sin^3α- 3sin^2α+ 3 sin^4 α ) - 3(1 + sin^4α- 2sin^2α) =3sin^4α - 2sin^6α-1$$
$$2 cos^6 α - 3 cos^4 α=2sin^4α - 3sin^6α-1$$
$$2 u_6 - 3 u_4 +1= (2cos^6 α + 2sin^6 α) - (3cos^4 α + 3sin^4 α) +1= 3sin^4α - 2sin^6α - 3sin^4α + 2sin^6α-1+1 $$
$$2 u_6 - 3 u_4 = 0$$
Hence Option D is the correct answer.