If cos^3 theta + cos^3 left(frac{2 pi}{3} + t | vedclass

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(B) Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos^3 x = \frac{\cos 3x + 3 \cos x}{4}$.Applying this to each term:$\frac{\cos 3

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$\frac{3}{4}$

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(B) Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos^3 x = \frac{\cos 3x + 3 \cos x}{4}$.Applying this to each term:$\frac{\cos 3 	heta + 3 \cos 	heta}{4} + \frac{\cos(2 \pi + 3 	heta) + 3 \cos(\frac{2 \pi}{3} + 	heta)}{4} + \frac{\cos(4 \pi + 3 	heta) + 3 \cos(\frac{4 \pi}{3} + 	heta)}{4} = \alpha \cos 3 	heta$Since $\cos(2 \pi + 3 	heta) = \cos 3 	heta$ and $\cos(4 \pi + 3 	heta) = \cos 3 	heta$,the equation becomes:$\frac{3 \cos 3 	heta + 3 \cos 	heta + 3 \cos(\frac{2 \pi}{3} + 	heta) + 3 \cos(\frac{4 \pi}{3} + 	heta)}{4} = \alpha \cos 3 	heta$Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$\cos(\frac{2 \pi}{3} + 	heta) + \cos(\frac{4 \pi}{3} + 	heta) = 2 \cos(\pi + 	heta) \cos(-\frac{\pi}{3}) = 2(-\cos 	heta)(\frac{1}{2}) = -\cos 	heta$Substituting this back:$\frac{3 \cos 3 	heta + 3 \cos 	heta + 3(-\cos 	heta)}{4} = \alpha \cos 3 	heta$$\frac{3 \cos 3 	heta}{4} = \alpha \cos 3 	heta$Thus,$\alpha = \frac{3}{4}$.

If $\cos^3 \theta + \cos^3 \left(\frac{2 \pi}{3} + \theta\right) + \cos^3 \left(\frac{4 \pi}{3} + \theta\right) = \alpha \cos 3 \theta$,then $\alpha$ is equal to

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