Let

$x= cos\theta + isin\theta \dfrac{1}{x}= cos\theta -isin\theta $

$x+\frac{1}{x}= 2 cos\theta $    $x+\dfrac{1}{x}= 2isin\theta$

$(2cos\theta) ^6 = (x+\dfrac{1}{x})^6 =x^6+6x^5.\dfrac{1}{x}+15x^4\dfrac{1}{x^2}+20x^3.\dfrac{1}{x^3}+15x^2.\dfrac{1}{x^4}+6x.\dfrac{1}{x^5}+\dfrac{1}{x^6}$ (by  binomial teorem

$=x^6 +6x^4+15x^2+20+15.\dfrac{1}{x^2}+6\dfrac{1}{x^4}+\dfrac{1}{x^6}---------(1) $

$Similarly(2isin\theta)^6= (x-\dfrac{1}{x})^6 =x^6-6x^4+15x^2-20+15.\dfrac{1}{x^2}-6.\dfrac{1}{x^4}+\dfrac{1}{x^6}$

$(i)^6 = (i^2)^3 = (-1)^3=-1$

   $\therefore (2sin\theta)^6 =-x^6+6x^4-15x^3+20-15\dfrac{1}{x^2}+6.\dfrac{1}{x^4}-\dfrac{1}{x^6} ------(2) $

Adding (1) and (2)

$(2cos\theta)^6 + (2sin\theta)^6 =12x^4+40+12\dfrac{1}{x^4}$

$2^6(cos^6\theta+sin^6\theta) = 12(x^4 + \dfrac{1}{x^4}) + 40 $

$(x^4 +\dfrac{1}{x^4}) =2 Cos 4\theta $

By De Moviie theorem

$Cos^6\theta + sin^6\theta = \dfrac{1}{2^6} (12(2cos4\theta) +40)$

$cos^6\theta + sin^6 \theta = \frac{3}{8} Cos4\theta +\frac{5}{8}$   Compare this with the problem statemet

$cos^6\theta +sin^6\theta = \alpha cos4\theta +\beta$

$\alpha = \dfrac{3}{8} \beta=\frac{5}{8}$

$\alpha +\beta = \dfrac{3}{8} + \dfrac{5}{8} = 1$

$\therefore \alpha + \beta = 1 ( proved)$

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