If x + frac{1}{x} = 2 cos theta, then x^3 + f | vedclass

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(B) Given $x + \frac{1}{x} = 2 \cos 	heta$. We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Substituting $a = x$ and $b = \frac{

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$2 \cos 3	heta$

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(B) Given $x + \frac{1}{x} = 2 \cos 	heta$. We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Substituting $a = x$ and $b = \frac{1}{x}$,we get: $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}ight)^3 - 3 \left(x \cdot \frac{1}{x}ight) \left(x + \frac{1}{x}ight)$ $= (2 \cos 	heta)^3 - 3(1)(2 \cos 	heta)$ $= 8 \cos^3 	heta - 6 \cos 	heta$ $= 2(4 \cos^3 	heta - 3 \cos 	heta)$ Using the trigonometric identity $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta$,we get: $= 2 \cos 3	heta$.

If $x + \frac{1}{x} = 2 \cos \theta,$ then $x^3 + \frac{1}{x^3} = $

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