$$sin^{6}θ + cos^{6} θ$$ is equal to

$$sin^6 θ + cos^6 θ = (sin^2θ)^3 + (cos^2θ)^3 $$
Consider
$$(sin^2 θ + cos^2 θ)^3 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1= sin^6 θ + cos^6 θ + 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) $$
$$1 - 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) = sin^6 θ + cos^6 θ $$
$$1 - 3 sin^2 θ cos^2 θ = sin^6 θ + cos^6 θ $$
Hence Option B is the correct answer.