Let $\cos θ+ i\sin θ=e^{iθ}=x $

$$∴\cos θ-i \sin θ=e^{-iθ}=\dfrac 1x $$

$ ∴\cos θ= \dfrac {e^{iθ} + e^{-iθ}}2= (\dfrac {x+\dfrac 1x}2) ………(1) \\ \text {Similarly } ∴ \sin θ= \dfrac {e^{iθ} + e^{-iθ}}{2!}= (\dfrac {x-\dfrac 1x}{2!}) ………(2) \\ \text { Now } \cos^6 θ=\dfrac 1{2^6} (x + \dfrac 1x)^6 \text {from } (1) \\ \cos^6⁡ θ=\dfrac 1{64}(x^6+ 6x^5.\dfrac 1x + 15x^4.\dfrac1{x^2} + 20x^3.\dfrac1{x^3} + 15x^2.\dfrac 1{x^4} +6x. \dfrac 1{x^5} + \dfrac 1{x^6} ) \\ \cos^6 θ= \dfrac 1{64} (x^6+6x^4 + 15x^2 + 20 + 15. \dfrac 1{x^2} + 6.\dfrac 1{x^4} + \dfrac 1{x^6} ) …… (3) \\ \sin^6⁡ θ= \dfrac 1{(2i)^6} (x - \dfrac 1x)^6 \text { from }(2) \\ = \dfrac 1{64(i)^6} [x^6 - 6x^5.\dfrac 1x + 15x^4.\dfrac 1{x^2} -20x^3.\dfrac 1{x^3} + 15x^2. \dfrac 1{x^4} - 6x.\dfrac 1{x^5} + \dfrac 1{x^6} ] \\ \sin^6 θ= \dfrac {-1}{64}[x^6 - 6x^4 + 15x^2-20 + 15.\dfrac 1{x^2} -6. \dfrac 1{x^4} + \dfrac 1{x^6} ] \\ \sin^6 θ=\dfrac 1{64} [-x^6+6x^4 - 15x^2 + 20-15. \dfrac 1{x^2} + 6.\dfrac 1{x^4} - \dfrac 1{x^6} ] ……. (4) $

Now adding (3) and (4) we get

$\cos^6 θ + \sin^6 θ= \dfrac 1{64} (x^6+6x^4+15x^2+20+15.\dfrac 1{x^2} +6.\dfrac 1{x^4} +\dfrac 1{x^6} +6.\dfrac 1{x^4} + \dfrac 1{x^6} - x^6 + 6x^4- 15x^2+20- \dfrac {15}{x^2} + \dfrac 6{x^4} - \dfrac 1{x^6} ) \\ =\dfrac 1{64} [12x^4+40 + \dfrac {12}{x^4} ] \\ = \dfrac 1{64} [12(x^4 + \dfrac 1{x^4} )+40] \\ ∴ (x^4 + \dfrac 1{x^4} ) = e^{4iθ} + e^{-4iθ} \\ = 2\cos4θ \\ \cos^6 θ + \sin^6 θ= \dfrac 1{64} [12×2 \cos4 θ+40] \\ = \dfrac 1{64} 8[3\cos4θ+5] \\ = \dfrac 18(3\cosθ+5) $

Hence proved.